Samacheer Kalvi Class 12 Maths Solutions Chapter 9 Applications of Integration Exercise 9.10

Choose the Most Suitable Answer From the Given Four Alternatives:

Question 1. The value of \( \int_{0}^{2/3} \frac { dx }{ \sqrt{4-9x^2} } \)
(a) \( \frac { \pi }{ 6 } \)
(b) \( \frac { \pi }{ 2 } \)
(c) \( \frac { \pi }{ 4 } \)
(d) \( \frac { \pi }{ 3 } \)
Answer: (a) \( \frac { \pi }{ 6 } \)
This integral can be solved by recognizing it as a form of the inverse sine function. First, we need to adjust the denominator to match the standard integral form of \( \frac{1}{\sqrt{a^2 - x^2}} \). Then, apply the limits of integration to find the definite value.
Here's the detailed working:
\( I = \int_{0}^{2/3} \frac { dx }{ \sqrt{4-9x^2} } \)
We can rewrite \( \sqrt{4-9x^2} \) as \( \sqrt{9(\frac{4}{9} - x^2)} = 3\sqrt{(\frac{2}{3})^2 - x^2} \).
So, \( I = \int_{0}^{2/3} \frac { dx }{ 3\sqrt{(\frac{2}{3})^2 - x^2} } \)
\( I = \frac{1}{3} \int_{0}^{2/3} \frac { dx }{ \sqrt{(\frac{2}{3})^2 - x^2} } \)
This is in the form \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) \). Here, \( a = \frac{2}{3} \).
So, \( I = \frac{1}{3} \left[ \sin^{-1}\left(\frac{x}{2/3}\right) \right]_{0}^{2/3} \)
\( I = \frac{1}{3} \left[ \sin^{-1}\left(\frac{3x}{2}\right) \right]_{0}^{2/3} \)
Now, apply the limits:
\( I = \frac{1}{3} \left[ \sin^{-1}\left(\frac{3(2/3)}{2}\right) - \sin^{-1}\left(\frac{3(0)}{2}\right) \right] \)
\( I = \frac{1}{3} \left[ \sin^{-1}(1) - \sin^{-1}(0) \right] \)
We know \( \sin^{-1}(1) = \frac{\pi}{2} \) and \( \sin^{-1}(0) = 0 \).
\( I = \frac{1}{3} \left[ \frac{\pi}{2} - 0 \right] \)
\( I = \frac{1}{3} \times \frac{\pi}{2} \)
\( I = \frac{\pi}{6} \)
In simple words: To solve this, change the bottom part of the fraction to match a known integral pattern for inverse sine. Then, put in the start and end values to find the final answer. The answer is pi divided by six.

🎯 Exam Tip: Always look for standard integral forms like \( \sin^{-1}(x/a) \) or \( \tan^{-1}(x/a) \) after simplifying the expression. Remember the values of inverse trigonometric functions at common angles like 0 and 1.

Question 2. The value of \( \int_{-1}^{2} |x| dx \)
(a) \( \frac { 1 }{ 2 } \)
(b) \( \frac { 3 }{ 2 } \)
(c) \( \frac { 5 }{ 2 } \)
(d) \( \frac { 7 }{ 2 } \)
Answer: (c) \( \frac { 5 }{ 2 } \)
When integrating a function with an absolute value like \( |x| \), it's important to split the integral at the point where the expression inside the absolute value changes sign. For \( |x| \), this point is \( x=0 \). Therefore, we separate the integral into two parts: one for \( x < 0 \) where \( |x| = -x \), and one for \( x \ge 0 \) where \( |x| = x \). This allows for a correct evaluation of the area under the curve.
Here's the detailed working:
The absolute value function \( |x| \) is defined as:
\( |x| = -x \) if \( x < 0 \)
\( |x| = x \) if \( x \ge 0 \)
We need to evaluate \( \int_{-1}^{2} |x| dx \). We split the integral at \( x=0 \).
\( \int_{-1}^{2} |x| dx = \int_{-1}^{0} -x dx + \int_{0}^{2} x dx \)
Now, we integrate each part:
\( \int_{-1}^{0} -x dx = \left[ -\frac{x^2}{2} \right]_{-1}^{0} \)
\( = \left( -\frac{0^2}{2} \right) - \left( -\frac{(-1)^2}{2} \right) \)
\( = 0 - \left( -\frac{1}{2} \right) \)
\( = \frac{1}{2} \)
And,
\( \int_{0}^{2} x dx = \left[ \frac{x^2}{2} \right]_{0}^{2} \)
\( = \left( \frac{2^2}{2} \right) - \left( \frac{0^2}{2} \right) \)
\( = \frac{4}{2} - 0 \)
\( = 2 \)
Finally, add the results from both parts:
\( \int_{-1}^{2} |x| dx = \frac{1}{2} + 2 \)
\( = \frac{1}{2} + \frac{4}{2} \)
\( = \frac{5}{2} \)
In simple words: When you see \( |x| \) in an integral, break the integral into two parts: one where x is negative (so \( |x| \) becomes \( -x \)) and one where x is positive (so \( |x| \) stays \( x \)). Then, do each integral and add the answers.

🎯 Exam Tip: Always split integrals involving absolute values at the points where the argument of the absolute value becomes zero. This ensures you apply the correct function definition for each interval.

Question 3. For any value of \( n \in Z \int_{0}^{\pi} e^{\cos^2x} \cos^3 [(2n + 1) x] dx \)
(a) \( \frac { \pi }{ 2 } \)
(b) \( \pi \)
(c) \( 0 \)
(d) \( 2 \)
Answer: (c) \( 0 \)
This integral can be solved using a property of definite integrals: if \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \). We'll apply this to the given function and then combine the original and modified integrals. This is a powerful technique for integrals with symmetric limits.
Here's the detailed working:
Let \( I = \int_{0}^{\pi} e^{\cos^2x} \cos^3 [(2n+1)x] dx \) .....(1)
Use the property: \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \)
Here, \( a = \pi \). So we replace \( x \) with \( \pi - x \).
\( \cos^2(\pi-x) = (-\cos x)^2 = \cos^2x \)
\( \cos^3[(2n+1)(\pi-x)] = \cos^3[(2n+1)\pi - (2n+1)x] \)
Since \( (2n+1)\pi \) is an odd multiple of \( \pi \), \( \cos((2n+1)\pi - \theta) = -\cos \theta \).
So, \( \cos^3[(2n+1)\pi - (2n+1)x] = (-\cos[(2n+1)x])^3 = -\cos^3[(2n+1)x] \)
Therefore, substituting \( \pi - x \) into the integral:
\( I = \int_{0}^{\pi} e^{\cos^2(\pi-x)} \cos^3 [(2n+1)(\pi-x)] dx \)
\( I = \int_{0}^{\pi} e^{\cos^2x} (-\cos^3 [(2n+1)x]) dx \)
\( I = - \int_{0}^{\pi} e^{\cos^2x} \cos^3 [(2n+1)x] dx \) .....(2)
Notice that equation (2) is \( I = -I \).
Adding (1) and (2):
\( I + I = \int_{0}^{\pi} e^{\cos^2x} \cos^3 [(2n+1)x] dx + \left( - \int_{0}^{\pi} e^{\cos^2x} \cos^3 [(2n+1)x] dx \right) \)
\( 2I = 0 \)
\( I = 0 \)
In simple words: This integral is solved by a trick: imagine adding the original integral to a version where you've replaced x with (pi - x). Because of how cosine works with pi, the new integral will be the negative of the old one, making them cancel out to zero when added together.

🎯 Exam Tip: For definite integrals with limits from 0 to 'a', always consider using the property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \). This is especially helpful when one part of the integrand changes sign with this substitution, leading to cancellation.

Question 4. The value of \( \int_{-\pi/2}^{\pi/2} \sin^2 x \cos x dx \)
(a) \( \frac { 3 }{ 2 } \)
(b) \( \frac { 1 }{ 2 } \)
(c) \( 0 \)
(d) \( \frac { 2 }{ 3 } \)
Answer: (d) \( \frac { 2 }{ 3 } \)
This integral has symmetric limits, from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \). For such integrals, it's useful to check if the function is even or odd. If the function \( f(x) \) is even (meaning \( f(-x) = f(x) \)), then \( \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx \). If it's odd (meaning \( f(-x) = -f(x) \)), then the integral is 0. Once we confirm it's an even function, we can simplify the integral and use substitution.
Here's the detailed working:
Let \( f(x) = \sin^2 x \cos x \).
Check if \( f(x) \) is an even or odd function:
\( f(-x) = \sin^2 (-x) \cos (-x) \)
Since \( \sin(-x) = -\sin x \) and \( \cos(-x) = \cos x \):
\( f(-x) = (-\sin x)^2 (\cos x) = \sin^2 x \cos x \)
So, \( f(-x) = f(x) \), which means \( f(x) \) is an even function.
For an even function with symmetric limits \( \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx \).
Therefore, \( \int_{-\pi/2}^{\pi/2} \sin^2 x \cos x dx = 2 \int_{0}^{\pi/2} \sin^2 x \cos x dx \)
Now, use substitution. Let \( t = \sin x \).
Then \( dt = \cos x dx \).
Change the limits of integration:
When \( x = 0 \), \( t = \sin 0 = 0 \).
When \( x = \frac{\pi}{2} \), \( t = \sin \frac{\pi}{2} = 1 \).
The integral becomes:
\( 2 \int_{0}^{1} t^2 dt \)
\( = 2 \left[ \frac{t^3}{3} \right]_{0}^{1} \)
\( = 2 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) \)
\( = 2 \left( \frac{1}{3} - 0 \right) \)
\( = 2 \times \frac{1}{3} \)
\( = \frac{2}{3} \)
In simple words: First, check if the function inside the integral is "even" (meaning it looks the same when you flip it over the y-axis). If it is, you can just calculate from 0 to the upper limit and double the result. Then, use a simple substitution to make the integral easier to solve.

🎯 Exam Tip: Always test for even or odd functions when the integral limits are symmetric (e.g., from -a to a). This can often simplify the problem significantly, sometimes even making the integral zero immediately.

Question 5. The value of \( \int_{-4}^{4}\left[\tan^{-1}\left(\frac { x^2 }{ x^4+1 }\right) + \tan^{-1}\left(\frac {x^4+1 }{ x^2 }\right)\right] dx \)
(a) \( \pi \)
(b) \( 2\pi \)
(c) \( 3\pi \)
(d) \( 4\pi \)
Answer: (d) \( 4\pi \)
This integral uses a very useful property of inverse trigonometric functions: \( \tan^{-1}(y) + \cot^{-1}(y) = \frac{\pi}{2} \). Since \( \tan^{-1}(y) + \tan^{-1}(\frac{1}{y}) = \frac{\pi}{2} \) for \( y > 0 \), we can simplify the expression inside the integral. The terms \( \frac{x^2}{x^4+1} \) and \( \frac{x^4+1}{x^2} \) are reciprocals of each other, and for the given limits, \( x^2 \) is always positive, allowing us to apply this property.
Here's the detailed working:
Let \( I = \int_{-4}^{4}\left[\tan^{-1}\left(\frac { x^2 }{ x^4+1 }\right) + \tan^{-1}\left(\frac {x^4+1 }{ x^2 }\right)\right] dx \)
Let \( y = \frac{x^2}{x^4+1} \). Then \( \frac{1}{y} = \frac{x^4+1}{x^2} \).
We know the identity \( \tan^{-1}(y) + \tan^{-1}\left(\frac{1}{y}\right) = \frac{\pi}{2} \) for \( y > 0 \).
Since \( x^2 > 0 \) for \( x \ne 0 \), and \( x^4+1 > 0 \), it follows that \( y = \frac{x^2}{x^4+1} > 0 \).
So, the expression inside the integral simplifies to \( \frac{\pi}{2} \).
\( I = \int_{-4}^{4} \frac{\pi}{2} dx \)
Since \( \frac{\pi}{2} \) is a constant, we can take it out of the integral:
\( I = \frac{\pi}{2} \int_{-4}^{4} dx \)
\( I = \frac{\pi}{2} [x]_{-4}^{4} \)
\( I = \frac{\pi}{2} [4 - (-4)] \)
\( I = \frac{\pi}{2} [4 + 4] \)
\( I = \frac{\pi}{2} \times 8 \)
\( I = 4\pi \)
In simple words: The two inverse tangent parts inside the integral are special because they are linked by a rule: if you have `tan⁻¹(y)` plus `tan⁻¹(1/y)`, the answer is always `pi/2`. Since `x²/(x⁴+1)` and `(x⁴+1)/x²` are exact opposites, their sum becomes `pi/2`. Then, you just integrate `pi/2` from -4 to 4.

🎯 Exam Tip: Recognize reciprocal forms within inverse tangent functions. The property \( \tan^{-1}(A) + \tan^{-1}(1/A) = \pi/2 \) (for \( A>0 \)) is a frequent shortcut in integration problems.

Question 6. The value of \( \int_{-\pi/4}^{\pi/4} \left(\frac { 2x^7-3x^5+7x^3-x+1 }{ \cos^2x }\right) dx \) is
(a) \( 4 \)
(b) \( 3 \)
(c) \( 2 \)
(d) \( 0 \)
Answer: (c) \( 2 \)
This integral has symmetric limits, so we should separate the integrand into odd and even functions. Recall that for an odd function \( f(x) \) (where \( f(-x) = -f(x) \)), \( \int_{-a}^{a} f(x) dx = 0 \). For an even function \( g(x) \) (where \( g(-x) = g(x) \)), \( \int_{-a}^{a} g(x) dx = 2 \int_{0}^{a} g(x) dx \). We will split the numerator and apply this property to each term.
Here's the detailed working:
Let the integral be \( I \).
\( I = \int_{-\pi/4}^{\pi/4} \frac{2x^7-3x^5+7x^3-x+1}{\cos^2x} dx \)
We can split the integral based on the terms in the numerator:
\( I = \int_{-\pi/4}^{\pi/4} \left( \frac{2x^7}{\cos^2x} - \frac{3x^5}{\cos^2x} + \frac{7x^3}{\cos^2x} - \frac{x}{\cos^2x} + \frac{1}{\cos^2x} \right) dx \)
Let's examine each term's parity (even or odd function):
For a term \( \frac{x^k}{\cos^2x} \): \( \cos^2(-x) = (\cos x)^2 = \cos^2x \), so the denominator is always even.
If \( k \) is odd, then \( (-x)^k = -x^k \), making the term \( \frac{-x^k}{\cos^2x} \), which is an odd function.
If \( k \) is even, then \( (-x)^k = x^k \), making the term \( \frac{x^k}{\cos^2x} \), which is an even function.

\( \frac{2x^7}{\cos^2x} \) is odd, so \( \int_{-\pi/4}^{\pi/4} \frac{2x^7}{\cos^2x} dx = 0 \).
\( \frac{-3x^5}{\cos^2x} \) is odd, so \( \int_{-\pi/4}^{\pi/4} \frac{-3x^5}{\cos^2x} dx = 0 \).
\( \frac{7x^3}{\cos^2x} \) is odd, so \( \int_{-\pi/4}^{\pi/4} \frac{7x^3}{\cos^2x} dx = 0 \).
\( \frac{-x}{\cos^2x} \) is odd, so \( \int_{-\pi/4}^{\pi/4} \frac{-x}{\cos^2x} dx = 0 \).
\( \frac{1}{\cos^2x} = \sec^2x \). This is an even function (since \( \sec^2(-x) = \sec^2x \)).
So, \( \int_{-\pi/4}^{\pi/4} \frac{1}{\cos^2x} dx = 2 \int_{0}^{\pi/4} \sec^2x dx \).

Therefore, the original integral simplifies to:
\( I = 0 - 0 + 0 - 0 + 2 \int_{0}^{\pi/4} \sec^2x dx \)
\( I = 2 \int_{0}^{\pi/4} \sec^2x dx \)
The integral of \( \sec^2x \) is \( \tan x \).
\( I = 2 [\tan x]_{0}^{\pi/4} \)
\( I = 2 (\tan(\pi/4) - \tan(0)) \)
\( I = 2 (1 - 0) \)
\( I = 2 \times 1 \)
\( I = 2 \)
In simple words: Look at each part of the fraction inside the integral. If a part behaves like an "odd" function (meaning it flips sign if you replace x with -x), its integral over symmetric limits is zero. The remaining "even" part (which stays the same when x becomes -x) can be integrated from 0 to the upper limit and then doubled. Only the `1/cos²x` part is even, which simplifies to `2 * tan(x)` from 0 to pi/4.

🎯 Exam Tip: For integrals with symmetric limits, always check the parity of the integrand. Identifying odd functions can drastically simplify the problem by immediately reducing their contribution to zero.

Question 7. If \( f(x) = \int_{0}^{x} \cos t dt \), then \( \frac { df }{ dx } = \)
(a) \( \cos x - x \sin x \)
(b) \( \sin x + x \cos x \)
(c) \( x \cos x \)
(d) \( x \sin x \)
Answer: (c) \( x \cos x \)
This problem involves differentiating an integral with a variable upper limit. This is a direct application of the Fundamental Theorem of Calculus (Leibniz integral rule). If \( F(x) = \int_{a}^{x} g(t) dt \), then \( F'(x) = g(x) \). However, the integrand here is \( t \cos t \), not just \( \cos t \). So the problem statement seems to have a typo, and should likely be \( f(x) = \int_{0}^{x} t \cos t dt \), based on the solution provided. Assuming the integrand is \( t \cos t \), we would integrate by parts first, and then differentiate using the Fundamental Theorem of Calculus.
Here's the detailed working:
Let's assume the question meant \( f(x) = \int_{0}^{x} t \cos t dt \). (Based on common problems where `x cos x` is the derivative).
First, evaluate the integral \( \int t \cos t dt \) using integration by parts, \( \int u dv = uv - \int v du \).
Let \( u = t \), then \( du = dt \).
Let \( dv = \cos t dt \), then \( v = \int \cos t dt = \sin t \).
So, \( \int t \cos t dt = t \sin t - \int \sin t dt \)
\( = t \sin t - (-\cos t) \)
\( = t \sin t + \cos t \)
Now, apply the limits from 0 to \( x \):
\( f(x) = [t \sin t + \cos t]_{0}^{x} \)
\( f(x) = (x \sin x + \cos x) - (0 \cdot \sin 0 + \cos 0) \)
\( f(x) = x \sin x + \cos x - (0 + 1) \)
\( f(x) = x \sin x + \cos x - 1 \)
Now, differentiate \( f(x) \) with respect to \( x \):
\( \frac{df}{dx} = \frac{d}{dx} (x \sin x + \cos x - 1) \)
Use the product rule for \( x \sin x \): \( \frac{d}{dx}(uv) = u'v + uv' \).
\( \frac{d}{dx}(x \sin x) = (1)(\sin x) + (x)(\cos x) = \sin x + x \cos x \)
\( \frac{d}{dx}(\cos x) = -\sin x \)
\( \frac{d}{dx}(1) = 0 \)
So, \( \frac{df}{dx} = (\sin x + x \cos x) - \sin x - 0 \)
\( \frac{df}{dx} = x \cos x \)
This matches option (c). This derivation relies on the assumption that the integrand in the question was \( t \cos t \), not \( \cos t \). If it were just \( \int_{0}^{x} \cos t dt \), then \( f(x) = [\sin t]_{0}^{x} = \sin x - \sin 0 = \sin x \), and \( \frac{df}{dx} = \cos x \), which is not an option. Thus, the implicit understanding is that the integrand includes \( t \).
In simple words: This question asks you to first integrate a function involving `t cos t` and then differentiate the result. You use a method called 'integration by parts' to solve the integral. After you get the new function, you take its derivative. The number 'x' shows up in the answer because it was the upper limit of the integral.

🎯 Exam Tip: Be careful with the variables when using the Fundamental Theorem of Calculus. If the integral is \( \int_{a}^{x} g(t) dt \), its derivative with respect to \( x \) is simply \( g(x) \). However, if \( g(t) \) itself involves \( t \) (like \( t \cos t \)), you must first evaluate the integral before differentiating, or apply the Leibniz rule more generally if the limits are also functions of \( x \).

Question 8. The area between \( y^2 = 4x \) and its latus rectum is
(a) \( \frac { 2 }{ 3 } \)
(b) \( \frac { 4 }{ 3 } \)
(c) \( \frac { 8 }{ 3 } \)
(d) \( \frac { 5 }{ 3 } \)
Answer: (c) \( \frac { 8 }{ 3 } \)
The equation \( y^2 = 4x \) represents a parabola opening to the right with its vertex at the origin. The latus rectum of a parabola \( y^2 = 4ax \) is the line \( x=a \). For \( y^2 = 4x \), we have \( 4a=4 \), so \( a=1 \). Thus, the latus rectum is the line \( x=1 \). We need to find the area enclosed by the parabola and the line \( x=1 \). Since the parabola is symmetric about the x-axis, we can find the area in the first quadrant and double it.
Here's the detailed working:
The equation of the parabola is \( y^2 = 4x \).
The standard form is \( y^2 = 4ax \), so \( 4a = 4 \implies a = 1 \).
The latus rectum is the line \( x = a \), which means \( x = 1 \).
To find the area, we integrate \( y \) with respect to \( x \). From \( y^2 = 4x \), we get \( y = \pm 2\sqrt{x} \).
We are interested in the area bounded by the parabola and the line \( x=1 \), from \( x=0 \) to \( x=1 \).
Since the parabola is symmetric about the x-axis, the total area is twice the area in the first quadrant.
Area \( = 2 \int_{0}^{1} y dx \)
Area \( = 2 \int_{0}^{1} 2\sqrt{x} dx \)
Area \( = 4 \int_{0}^{1} x^{1/2} dx \)
Now, integrate \( x^{1/2} \):
Area \( = 4 \left[ \frac{x^{1/2+1}}{1/2+1} \right]_{0}^{1} \)
Area \( = 4 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} \)
Area \( = 4 \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} \)
Area \( = \frac{8}{3} [x^{3/2}]_{0}^{1} \)
Now, apply the limits:
Area \( = \frac{8}{3} (1^{3/2} - 0^{3/2}) \)
Area \( = \frac{8}{3} (1 - 0) \)
Area \( = \frac{8}{3} \) square units.

In simple words: This problem asks for the space between a parabola (a U-shaped curve) and a special line called its 'latus rectum'. You first find out where the latus rectum is for this specific parabola. Then, you use a math tool called integration to measure the area. Because the curve is even on both sides, you calculate one half and double it to get the total area.

🎯 Exam Tip: Always sketch the curves involved in area problems. Identifying symmetry (like for parabolas) can simplify the calculation by allowing you to integrate over half the region and then multiply by two.

Question 9. The value of \( \int_{0}^{1} x (1-x)^{99} dx \) is
(a) \( \frac { 1 }{ 11000 } \)
(b) \( \frac { 1 }{ 10100 } \)
(c) \( \frac { 1 }{ 10000 } \)
(d) \( \frac { 1 }{ 10001 } \)
Answer: (b) \( \frac { 1 }{ 10100 } \)
This integral is in the form of a Beta function integral, which is defined as \( B(m, n) = \int_{0}^{1} x^{m-1} (1-x)^{n-1} dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} \). Using the property that \( \Gamma(z+1) = z! \) for integer \( z \), we can directly calculate the value. This method provides a shortcut for certain types of definite integrals.
Here's the detailed working:
The given integral is \( \int_{0}^{1} x (1-x)^{99} dx \).
Compare this with the Beta function integral form \( \int_{0}^{1} x^{m-1} (1-x)^{n-1} dx \).
For \( x^{m-1} \): we have \( x^1 \), so \( m-1 = 1 \implies m = 2 \).
For \( (1-x)^{n-1} \): we have \( (1-x)^{99} \), so \( n-1 = 99 \implies n = 100 \).
The integral can be evaluated as \( \frac{m! n!}{(m+n+1)!} \). This is actually \( \frac{\Gamma(m+1)\Gamma(n+1)}{\Gamma(m+n+2)} \).
A more direct formula (from the source's hint): \( \int_{0}^{1} x^m (1-x)^n dx = \frac{m! n!}{(m+n+1)!} \)
Here, \( m=1 \) and \( n=99 \).
So, the value of the integral is \( \frac{1! \times 99!}{(1+99+1)!} \)
\( = \frac{1! \times 99!}{101!} \)
\( = \frac{1 \times 99!}{101 \times 100 \times 99!} \)
\( = \frac{1}{101 \times 100} \)
\( = \frac{1}{10100} \)
In simple words: This integral follows a special pattern called the Beta function. You just need to match the powers of 'x' and '(1-x)' to find the right numbers, then plug them into a simple factorial formula. This saves a lot of time compared to normal integration.

🎯 Exam Tip: Memorize the Beta function integral form and its relation to factorials. It's a powerful shortcut for definite integrals from 0 to 1 involving products of powers of \( x \) and \( (1-x) \).

Question 10. The value of \( \int_{0}^{\pi} \frac { dx }{ 1+5^{\cos x} } \) is
(a) \( \frac { \pi }{ 2 } \)
(b) \( \pi \)
(c) \( \frac { 3\pi }{ 2 } \)
(d) \( 2\pi \)
Answer: (a) \( \frac { \pi }{ 2 } \)
This integral can be solved using the property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \). By applying this property, we can transform the integral into a new form that, when added to the original, simplifies significantly. This approach is particularly effective when the term involving \( a-x \) leads to a reciprocal relationship.
Here's the detailed working:
Let \( I = \int_{0}^{\pi} \frac { dx }{ 1+5^{\cos x} } \) .....(1)
Use the property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \). Here \( a = \pi \).
Replace \( x \) with \( \pi - x \):
\( I = \int_{0}^{\pi} \frac { dx }{ 1+5^{\cos(\pi - x)} } \)
Since \( \cos(\pi - x) = -\cos x \):
\( I = \int_{0}^{\pi} \frac { dx }{ 1+5^{-\cos x} } \)
Rewrite \( 5^{-\cos x} \) as \( \frac{1}{5^{\cos x}} \):
\( I = \int_{0}^{\pi} \frac { dx }{ 1+\frac{1}{5^{\cos x}} } \)
Simplify the denominator:
\( I = \int_{0}^{\pi} \frac { dx }{ \frac{5^{\cos x}+1}{5^{\cos x}} } \)
\( I = \int_{0}^{\pi} \frac { 5^{\cos x} }{ 5^{\cos x}+1 } dx \) .....(2)
Now, add equation (1) and equation (2):
\( 2I = \int_{0}^{\pi} \frac { dx }{ 1+5^{\cos x} } + \int_{0}^{\pi} \frac { 5^{\cos x} }{ 5^{\cos x}+1 } dx \)
Since the denominators are the same, we can combine the numerators:
\( 2I = \int_{0}^{\pi} \frac { 1 + 5^{\cos x} }{ 1 + 5^{\cos x} } dx \)
\( 2I = \int_{0}^{\pi} 1 dx \)
\( 2I = [x]_{0}^{\pi} \)
\( 2I = \pi - 0 \)
\( 2I = \pi \)
\( I = \frac{\pi}{2} \)
In simple words: This integral is solved by adding it to a modified version of itself. By replacing 'x' with 'pi - x', the integral changes in a way that when you add the original and modified integrals, many terms cancel out, making the integral very simple to solve.

🎯 Exam Tip: For integrals of the form \( \int_{0}^{a} f(x) dx \), always try using the property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \). This is especially useful when the function contains terms like \( \cos x \) or \( \sin x \) which change sign or value predictably with \( a-x \) substitution.

Question 11. If \( \frac { \Gamma(n+2) }{ \Gamma n } = 90 \) then \( n \) is
(a) \( 10 \)
(b) \( 5 \)
(c) \( 8 \)
(d) \( 9 \)
Answer: (d) \( 9 \)
This problem involves the Gamma function, which is a generalization of the factorial function to real and complex numbers. A key property of the Gamma function is \( \Gamma(z+1) = z\Gamma(z) \). By repeatedly applying this property, we can simplify the ratio of Gamma functions and solve for \( n \). This method is fundamental to working with Gamma functions.
Here's the detailed working:
We are given \( \frac{\Gamma(n+2)}{\Gamma n} = 90 \).
Use the property \( \Gamma(z+1) = z\Gamma(z) \).
Apply this to the numerator \( \Gamma(n+2) \):
\( \Gamma(n+2) = (n+1)\Gamma(n+1) \)
Now, apply the property again to \( \Gamma(n+1) \):
\( \Gamma(n+1) = n\Gamma(n) \)
Substitute this back into the expression for \( \Gamma(n+2) \):
\( \Gamma(n+2) = (n+1) [n\Gamma(n)] \)
\( \Gamma(n+2) = n(n+1)\Gamma(n) \)
Now, substitute this into the given equation:
\( \frac{n(n+1)\Gamma(n)}{\Gamma n} = 90 \)
The \( \Gamma(n) \) terms cancel out (assuming \( n \) is not a non-positive integer, which is true for natural numbers).
\( n(n+1) = 90 \)
Expand the equation:
\( n^2 + n = 90 \)
Rearrange into a quadratic equation:
\( n^2 + n - 90 = 0 \)
Factor the quadratic equation. We need two numbers that multiply to -90 and add to 1. These numbers are 10 and -9.
\( (n+10)(n-9) = 0 \)
This gives two possible solutions for \( n \):
\( n+10 = 0 \implies n = -10 \)
\( n-9 = 0 \implies n = 9 \)
Since \( n \) usually refers to a positive integer in such contexts (especially for factorials or Gamma functions to be well-defined in this simplified manner), we take the positive value.
Therefore, \( n = 9 \).
In simple words: The Gamma function is like a special factorial. You use a rule that says `Gamma(number + 1) = number * Gamma(number)` to simplify the fraction. This turns the problem into a simple equation, which you can then solve for 'n'.

🎯 Exam Tip: Remember the fundamental property of the Gamma function: \( \Gamma(z+1) = z\Gamma(z) \). This property is crucial for simplifying expressions involving ratios of Gamma functions and solving related equations.

Question 12. The value of \( \int_{0}^{\pi/6} \cos^3 3x dx \) is
(a) \( \frac { 2 }{ 3 } \)
(b) \( \frac { 1 }{ 10100 } \)
(c) \( \frac { 1 }{ 9 } \)
(d) \( \frac { 1 }{ 3 } \)
Answer: None of the options. The correct value is \( \frac{2}{9} \)
To solve this integral, we first use a trigonometric identity to reduce the power of \( \cos^3 \theta \). The identity \( \cos 3\theta = 4\cos^3 \theta - 3\cos \theta \) can be rearranged to express \( \cos^3 \theta \) in terms of \( \cos 3\theta \) and \( \cos \theta \). After the substitution, the integral becomes much simpler to evaluate. This method is effective for odd powers of sine and cosine.
Here's the detailed working:
Let \( I = \int_{0}^{\pi/6} \cos^3 3x dx \).
First, let's use a substitution to simplify the argument of the cosine function. Let \( u = 3x \).
Then \( du = 3 dx \implies dx = \frac{1}{3} du \).
Change the limits of integration:
When \( x = 0 \), \( u = 3(0) = 0 \).
When \( x = \frac{\pi}{6} \), \( u = 3\left(\frac{\pi}{6}\right) = \frac{\pi}{2} \).
So the integral becomes:
\( I = \int_{0}^{\pi/2} \cos^3 u \left(\frac{1}{3} du\right) \)
\( I = \frac{1}{3} \int_{0}^{\pi/2} \cos^3 u du \)
Now, use the trigonometric identity: \( \cos 3u = 4\cos^3 u - 3\cos u \).
Rearrange to solve for \( \cos^3 u \):
\( 4\cos^3 u = \cos 3u + 3\cos u \)
\( \cos^3 u = \frac{1}{4} (\cos 3u + 3\cos u) \)
Substitute this into the integral:
\( I = \frac{1}{3} \int_{0}^{\pi/2} \frac{1}{4} (\cos 3u + 3\cos u) du \)
\( I = \frac{1}{12} \int_{0}^{\pi/2} (\cos 3u + 3\cos u) du \)
Now, integrate term by term:
\( \int \cos 3u du = \frac{\sin 3u}{3} \)
\( \int 3\cos u du = 3\sin u \)
So, \( I = \frac{1}{12} \left[ \frac{\sin 3u}{3} + 3\sin u \right]_{0}^{\pi/2} \)
Apply the limits of integration:
\( I = \frac{1}{12} \left[ \left( \frac{\sin(3\pi/2)}{3} + 3\sin(\pi/2) \right) - \left( \frac{\sin(0)}{3} + 3\sin(0) \right) \right] \)
We know: \( \sin(3\pi/2) = -1 \), \( \sin(\pi/2) = 1 \), \( \sin(0) = 0 \).
\( I = \frac{1}{12} \left[ \left( \frac{-1}{3} + 3(1) \right) - (0 + 0) \right] \)
\( I = \frac{1}{12} \left[ -\frac{1}{3} + 3 \right] \)
\( I = \frac{1}{12} \left[ -\frac{1}{3} + \frac{9}{3} \right] \)
\( I = \frac{1}{12} \left[ \frac{8}{3} \right] \)
\( I = \frac{8}{36} \)
\( I = \frac{2}{9} \)
The calculated value is \( \frac{2}{9} \). Since this is not among the given options (a) \( \frac{2}{3} \), (b) \( \frac{1}{10100} \), (c) \( \frac{1}{9} \), (d) \( \frac{1}{3} \), we state that the correct value is \( \frac{2}{9} \).
In simple words: First, make the integral simpler by changing `3x` to `u`. Then, use a special rule for `cos³u` to break it into easier parts. Integrate each part and put in the upper and lower numbers to get the final answer. The actual calculation gives `2/9`.

🎯 Exam Tip: For odd powers of \( \sin x \) or \( \cos x \), use the identities \( \sin 3x = 3\sin x - 4\sin^3 x \) or \( \cos 3x = 4\cos^3 x - 3\cos x \) to reduce the power. This simplifies the integrand into terms that are easy to integrate.

Question 13. The value of \( \int_{0}^{\pi} \sin^4 x dx \)
(a) \( \frac { 3\pi }{ 10 } \)
(b) \( \frac { 3\pi }{ 8 } \)
(c) \( \frac { 3\pi }{ 4 } \)
(d) \( \frac { 3\pi }{ 2 } \)
Answer: (b) \( \frac { 3\pi }{ 8 } \)
This integral involves an even power of \( \sin x \) over the interval \( [0, \pi] \). We can use a reduction formula or the property that \( \int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx \) if \( f(2a-x) = f(x) \). For \( \sin^4 x \), since \( \sin^4(\pi-x) = (\sin(\pi-x))^4 = (\sin x)^4 = \sin^4 x \), we can write \( \int_{0}^{\pi} \sin^4 x dx = 2 \int_{0}^{\pi/2} \sin^4 x dx \). Then, Wallis' formula can be applied to evaluate \( \int_{0}^{\pi/2} \sin^n x dx \).
Here's the detailed working:
Let \( I = \int_{0}^{\pi} \sin^4 x dx \).
First, check the property \( f(a-x) = f(x) \). Here \( a = \pi \).
Let \( f(x) = \sin^4 x \). Then \( f(\pi-x) = \sin^4(\pi-x) = (\sin(\pi-x))^4 = (\sin x)^4 = \sin^4 x \).
Since \( f(\pi-x) = f(x) \), we can use the property \( \int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx \).
Here, \( 2a = \pi \), so \( a = \pi/2 \).
\( I = 2 \int_{0}^{\pi/2} \sin^4 x dx \)
Now, use Wallis' formula for \( \int_{0}^{\pi/2} \sin^n x dx \):
If \( n \) is even, \( \int_{0}^{\pi/2} \sin^n x dx = \frac{(n-1)!!}{n!!} \times \frac{\pi}{2} \)
Here, \( n = 4 \) (even).
\( \int_{0}^{\pi/2} \sin^4 x dx = \frac{(4-1)!!}{4!!} \times \frac{\pi}{2} = \frac{3!!}{4!!} \times \frac{\pi}{2} \)
\( 3!! = 3 \times 1 = 3 \)
\( 4!! = 4 \times 2 = 8 \)
So, \( \int_{0}^{\pi/2} \sin^4 x dx = \frac{3}{8} \times \frac{\pi}{2} = \frac{3\pi}{16} \)
Now, substitute this back into the expression for \( I \):
\( I = 2 \times \frac{3\pi}{16} \)
\( I = \frac{3\pi}{8} \)
In simple words: You first use a rule to cut the integral's range in half and double the result because the function is symmetric. Then, you use a special formula called Wallis' formula to quickly calculate the integral of `sin` to a power from 0 to pi/2.

🎯 Exam Tip: For integrals of even powers of \( \sin x \) or \( \cos x \) over \( [0, \pi] \) or \( [0, \pi/2] \), Wallis' formula is a significant time-saver. Remember to adjust the limits correctly if starting from \( [0, \pi] \).

Question 14. The value of \( \int_{0}^{\infty} e^{-3x} x^2 dx \)
(a) \( \frac { 7 }{ 27 } \)
(b) \( \frac { 5 }{ 27 } \)
(c) \( \frac { 4 }{ 27 } \)
(d) \( \frac { 2 }{ 27 } \)
Answer: (d) \( \frac { 2 }{ 27 } \)
This integral is a direct application of the Gamma function integral definition. The integral \( \int_{0}^{\infty} e^{-\alpha x} x^n dx \) is equal to \( \frac{n!}{\alpha^{n+1}} \) for \( n \) being a non-negative integer. By identifying \( \alpha \) and \( n \) from the given integral, we can directly compute its value using this formula. This property is very useful for solving certain types of improper integrals.
Here's the detailed working:
The given integral is \( \int_{0}^{\infty} e^{-3x} x^2 dx \).
This integral matches the form of the Gamma function integral: \( \int_{0}^{\infty} e^{-\alpha x} x^n dx = \frac{n!}{\alpha^{n+1}} \).
By comparing the given integral with the formula:
We have \( \alpha = 3 \).
We have \( n = 2 \).
Now, substitute these values into the formula:
\( \int_{0}^{\infty} e^{-3x} x^2 dx = \frac{2!}{3^{2+1}} \)
\( = \frac{2!}{3^3} \)
Calculate the factorial and the power:
\( 2! = 2 \times 1 = 2 \)
\( 3^3 = 3 \times 3 \times 3 = 27 \)
So, the value of the integral is:
\( = \frac{2}{27} \)
In simple words: This integral can be solved using a special rule related to the Gamma function. You just need to match the numbers in the integral to the formula, then calculate the factorial and power to get the final answer. It's a quick way to solve these kinds of integrals.

🎯 Exam Tip: Recognize the Gamma function integral \( \int_{0}^{\infty} e^{-\alpha x} x^n dx = \frac{n!}{\alpha^{n+1}} \). This formula is a powerful tool for quickly evaluating improper integrals of this specific form. Ensure you correctly identify \( n \) and \( \alpha \).

Question 15. If \( \int_{0}^{a} \frac { 1 }{ 4+x^2 } dx = \frac { \pi }{ 8 } \) then \( a \) is
(a) \( 4 \)
(b) \( 1 \)
(c) \( 3 \)
(d) \( 2 \)
Answer: (d) \( 2 \)
This problem requires evaluating a definite integral of the form \( \frac{1}{a^2+x^2} \) and then solving for an unknown limit. The integral of \( \frac{1}{a^2+x^2} dx \) is \( \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) \). We will evaluate the integral with the given limits and then set it equal to \( \frac{\pi}{8} \) to find the value of \( a \). This is a common application of inverse tangent integrals.
Here's the detailed working:
We are given \( \int_{0}^{a} \frac{1}{4+x^2} dx = \frac{\pi}{8} \).
The integral of \( \frac{1}{b^2+x^2} dx = \frac{1}{b} \tan^{-1}\left(\frac{x}{b}\right) + C \).
Here, \( b^2 = 4 \implies b = 2 \).
So, \( \int_{0}^{a} \frac{1}{4+x^2} dx = \left[ \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) \right]_{0}^{a} \)
Apply the limits of integration:
\( = \frac{1}{2} \tan^{-1}\left(\frac{a}{2}\right) - \frac{1}{2} \tan^{-1}\left(\frac{0}{2}\right) \)
\( = \frac{1}{2} \tan^{-1}\left(\frac{a}{2}\right) - \frac{1}{2} \tan^{-1}(0) \)
Since \( \tan^{-1}(0) = 0 \):
\( = \frac{1}{2} \tan^{-1}\left(\frac{a}{2}\right) - 0 \)
So, \( \frac{1}{2} \tan^{-1}\left(\frac{a}{2}\right) = \frac{\pi}{8} \)
Multiply both sides by 2:
\( \tan^{-1}\left(\frac{a}{2}\right) = \frac{2\pi}{8} \)
\( \tan^{-1}\left(\frac{a}{2}\right) = \frac{\pi}{4} \)
Now, take the tangent of both sides:
\( \frac{a}{2} = \tan\left(\frac{\pi}{4}\right) \)
We know \( \tan\left(\frac{\pi}{4}\right) = 1 \).
\( \frac{a}{2} = 1 \)
\( a = 2 \)
In simple words: First, you solve the integral using the inverse tangent formula. Then, you put in the numbers for the start and end of the integral. Finally, you set the whole thing equal to `pi/8` and solve for `a` by using the tangent function.

🎯 Exam Tip: Remember the standard integral form for \( \int \frac{1}{a^2+x^2} dx \). When solving for an unknown limit, carefully evaluate the inverse tangent values, especially for common angles like \( \pi/4 \) or \( 0 \).

Question 16. The volume of solid of revolution of the region bounded by \( y^2 = x(a - x) \) about the x-axis is
(a) \( \pi a^3 \)
(b) \( \frac { \pi a^3 }{ 4 } \)
(c) \( \frac { \pi a^3 }{ 5 } \)
(d) \( \frac { \pi a^3 }{ 6 } \)
Answer: (d) \( \frac { \pi a^3 }{ 6 } \)
To find the volume of a solid of revolution about the x-axis, we use the disk method, where the volume \( V = \pi \int_{x_1}^{x_2} y^2 dx \). First, we need to find the limits of integration by determining where the curve \( y^2 = x(a-x) \) intersects the x-axis (i.e., where \( y=0 \)). Then, we substitute \( y^2 \) into the formula and evaluate the integral. This method is crucial for calculating volumes of solids formed by rotating a curve.
Here's the detailed working:
The given equation is \( y^2 = x(a-x) \).
To find the limits of integration (where the region starts and ends along the x-axis), we set \( y=0 \):
\( 0 = x(a-x) \)
This gives \( x=0 \) or \( a-x=0 \implies x=a \).
So, the region is bounded from \( x=0 \) to \( x=a \).
The formula for the volume of revolution about the x-axis is:
\( V = \pi \int_{0}^{a} y^2 dx \)
Substitute \( y^2 = x(a-x) = ax - x^2 \):
\( V = \pi \int_{0}^{a} (ax - x^2) dx \)
Now, integrate term by term:
\( V = \pi \left[ \frac{ax^2}{2} - \frac{x^3}{3} \right]_{0}^{a} \)
Apply the limits of integration:
\( V = \pi \left[ \left( \frac{a(a)^2}{2} - \frac{(a)^3}{3} \right) - \left( \frac{a(0)^2}{2} - \frac{(0)^3}{3} \right) \right] \)
\( V = \pi \left[ \left( \frac{a^3}{2} - \frac{a^3}{3} \right) - (0 - 0) \right] \)
\( V = \pi \left[ \frac{a^3}{2} - \frac{a^3}{3} \right] \)
Combine the fractions:
\( V = \pi \left[ \frac{3a^3 - 2a^3}{6} \right] \)
\( V = \pi \left[ \frac{a^3}{6} \right] \)
\( V = \frac{\pi a^3}{6} \)
In simple words: To find the volume of a shape made by spinning a curve around the x-axis, you use a formula with `pi` and the integral of `y²`. First, find the start and end points of the curve on the x-axis. Then, put `y²` into the formula and calculate the integral.

🎯 Exam Tip: For volumes of revolution, ensure you correctly identify the axis of revolution and the limits of integration. Use the disk method (integrating \( \pi y^2 dx \)) for revolution around the x-axis and the shell method for more complex shapes or different axes.

Question 17. If \( f(x) = \int_{1}^{x} \frac { e^{\sin u} }{ u } du, x > 1 \) and \( \int_{1}^{3} \frac { e^{\sin u} }{ u } du = \frac { 1 }{ 2 } [f(a) – f(1)] \) then one of the possible value of \( a \) is
(a) \( 3 \)
(b) \( 6 \)
(c) \( 9 \)
(d) \( 5 \)
Answer: (c) \( 9 \)
This problem involves the definition of a function as an integral and then a substitution to relate different integrals. The given integral is a form of a logarithmic integral function. By setting \( u = x^2 \) (as suggested by the hint's context which often uses a change of variable like this), we can transform the integral and then compare it with the definition of \( f(x) \). This highlights the power of substitution in simplifying integral equations.
Here's the detailed working:
Given \( f(x) = \int_{1}^{x} \frac{e^{\sin u}}{u} du \).
Also given \( \int_{1}^{3} \frac{e^{\sin u}}{u} du = \frac{1}{2} [f(a) - f(1)] \).
From the definition of \( f(x) \), we can see that:
\( f(3) = \int_{1}^{3} \frac{e^{\sin u}}{u} du \)
And \( f(1) = \int_{1}^{1} \frac{e^{\sin u}}{u} du = 0 \) (since the upper and lower limits are the same).
Substitute these into the second given equation:
\( f(3) = \frac{1}{2} [f(a) - 0] \)
\( f(3) = \frac{1}{2} f(a) \)
Now, let's look at the "Consider I" part of the hint, which suggests a substitution. The hint is for `I = ∫(e^sin(x²)/x)dx`. It makes a substitution \( u = x^2 \). Let's follow this implied transformation.
If we were to relate \( f(x) \) to an integral with \( x^2 \) in the argument, we might have something like:
Let \( I = \int_{1}^{3} \frac{e^{\sin u}}{u} du \).
We have \( f(a) = \int_{1}^{a} \frac{e^{\sin u}}{u} du \).
So, \( \int_{1}^{3} \frac{e^{\sin u}}{u} du = f(3) \).
The equation becomes: \( f(3) = \frac{1}{2} [f(a) - f(1)] \).
Since \( f(1) = 0 \), we have \( f(3) = \frac{1}{2} f(a) \).
This means \( 2f(3) = f(a) \).
Let's consider the hint's substitution: Put \( u = x^2 \). Then \( du = 2x dx \). This is for transforming a different integral, not directly for the given \( f(x) \). The hint seems to be for another problem that might lead to a similar structure. However, the problem implies comparing the integral values.
We are given: \( \int_{1}^{3} \frac{e^{\sin u}}{u} du = \frac{1}{2} [f(a) - f(1)] \).
We know \( f(x) = \int_{1}^{x} \frac{e^{\sin u}}{u} du \).
So, \( f(3) = \int_{1}^{3} \frac{e^{\sin u}}{u} du \).
And \( f(1) = \int_{1}^{1} \frac{e^{\sin u}}{u} du = 0 \).
Substitute these into the given equation:
\( f(3) = \frac{1}{2} [f(a) - 0] \)
\( f(3) = \frac{1}{2} f(a) \)
This means \( f(a) = 2f(3) \).
This problem seems to be designed such that a specific substitution might convert \( f(a) \) into \( 2f(3) \).
Let's examine the integral `I` from the hint:
`I = ∫(e^sin(x²)/x)dx` from 1 to 3.
Substitute \( u = x^2 \). Then \( du = 2x dx \), so \( \frac{du}{2x} = dx \). Since \( x^2=u \), \( x = \sqrt{u} \). So \( dx = \frac{du}{2\sqrt{u}} \). This is not matching the hint's \( \frac{du}{2x} \).
The hint shows \( dx = \frac{du}{2x} \), so \( \frac{dx}{x} = \frac{du}{2x^2} = \frac{du}{2u} \).
So, if \( I = \int_{1}^{3} \frac{e^{\sin(x^2)}}{x} dx \), and \( u = x^2 \), then \( \frac{dx}{x} = \frac{du}{2u} \).
The limits change from \( x=1 \implies u=1^2=1 \) and \( x=3 \implies u=3^2=9 \).
So, \( I = \int_{1}^{9} e^{\sin u} \frac{du}{2u} = \frac{1}{2} \int_{1}^{9} \frac{e^{\sin u}}{u} du \).
This means \( I = \frac{1}{2} f(9) \).
Now compare with the initial integral \( \int_{1}^{3} \frac{e^{\sin u}}{u} du \), which is \( f(3) \).
So, \( f(3) = \frac{1}{2} f(9) \).
We were given: \( \int_{1}^{3} \frac{e^{\sin u}}{u} du = \frac{1}{2} [f(a) - f(1)] \).
Substitute \( f(3) \) on the left side and \( f(1)=0 \) on the right side:
\( f(3) = \frac{1}{2} [f(a) - 0] \)
\( f(3) = \frac{1}{2} f(a) \)
Comparing \( f(3) = \frac{1}{2} f(9) \) with \( f(3) = \frac{1}{2} f(a) \), we can conclude that \( f(a) = f(9) \).
Since \( f(x) \) is a strictly increasing function (because \( \frac{d}{dx}f(x) = \frac{e^{\sin x}}{x} > 0 \) for \( x>1 \)), it must be injective, meaning if \( f(a)=f(9) \), then \( a=9 \).
Therefore, \( a = 9 \).
In simple words: This problem connects different integrals of the same type. By understanding how the function `f(x)` is defined and by using a clever substitution as hinted, you can compare the given equation to the integral definition. This reveals that the unknown value `a` must be 9 for the equation to hold true.

🎯 Exam Tip: When dealing with functions defined by integrals, carefully apply the Fundamental Theorem of Calculus. For integral equations, look for substitutions that transform one integral into another, allowing for direct comparison of the function values.

Question 18. The value of \( \int_{0}^{1} (\sin^{-1} x)^2 dx \) is
(a) \( \frac { \pi^2 }{ 4 } - 1 \)
(b) \( \frac { \pi^2 }{ 4 } + 2 \)
(c) \( \frac { \pi^2 }{ 4 } + 1 \)
(d) \( \frac { \pi^2 }{ 4 } - 2 \)
Answer: (d) \( \frac { \pi^2 }{ 4 } - 2 \)
Hint:
We want to find \( I = \int_{0}^{1} (\sin^{-1} x)^2 dx \)
We use integration by parts with \( u = (\sin^{-1} x)^2 \) and \( dv = dx \).
So, \( du = 2 \sin^{-1} x \cdot \frac{1}{\sqrt{1-x^2}} dx \) and \( v = x \).
Applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\( I = [x (\sin^{-1} x)^2]_{0}^{1} - \int_{0}^{1} x \cdot 2 \sin^{-1} x \cdot \frac{1}{\sqrt{1-x^2}} dx \)
Next, we handle the new integral \( \int_{0}^{1} \frac{2x \sin^{-1} x}{\sqrt{1-x^2}} dx \).
Let \( u' = \sin^{-1} x \) and \( dv' = \frac{2x}{\sqrt{1-x^2}} dx \).
To find \( v' \), we integrate \( \int \frac{2x}{\sqrt{1-x^2}} dx \). Let \( t = 1-x^2 \), so \( dt = -2x dx \).
Then \( \int \frac{-dt}{\sqrt{t}} = - \int t^{-1/2} dt = - (2t^{1/2}) = -2\sqrt{1-x^2} \).
So, \( u' = \sin^{-1} x \) and \( v' = -2\sqrt{1-x^2} \).
Also, \( du' = \frac{1}{\sqrt{1-x^2}} dx \).
Applying integration by parts again for the second integral:
\( \int_{0}^{1} \frac{2x \sin^{-1} x}{\sqrt{1-x^2}} dx = [\sin^{-1} x (-2\sqrt{1-x^2})]_{0}^{1} - \int_{0}^{1} (-2\sqrt{1-x^2}) \frac{1}{\sqrt{1-x^2}} dx \)
\( = [-2\sqrt{1-x^2} \sin^{-1} x]_{0}^{1} + \int_{0}^{1} 2 dx \)
\( = [(-2\sqrt{1-1} \sin^{-1} 1) - (-2\sqrt{1-0} \sin^{-1} 0)] + [2x]_{0}^{1} \)
\( = [0 - 0] + [2(1) - 2(0)] = 2 \)
Now, substitute this back into the expression for \( I \):
\( I = [x (\sin^{-1} x)^2]_{0}^{1} - 2 \)
\( = [1 (\sin^{-1} 1)^2 - 0 (\sin^{-1} 0)^2] - 2 \)
\( = [(\frac{\pi}{2})^2 - 0] - 2 \)
\( = \frac{\pi^2}{4} - 2 \)
In simple words: To solve this, we used a method called integration by parts twice. We carefully broke down the integral into smaller, easier parts and then put them back together. The final answer is the value of pi squared divided by four, minus two.

🎯 Exam Tip: When dealing with integrals involving inverse trigonometric functions or powers, consider using integration by parts. Sometimes, it might require applying the method more than once.

Question 19. The value of \( \int_{0}^{a} (\sqrt{a^2-x^2})^3 dx \) is
(a) \( \frac { \pi a^2 }{ 16 } \)
(b) \( \frac { 3 \pi a^4 }{ 16 } \)
(c) \( \frac { 3 \pi a^2 }{ 8 } \)
(d) \( \frac { 3 \pi a^4 }{ 8 } \)
Answer: (b) \( \frac { 3 \pi a^4 }{ 16 } \)
Hint:
Let the integral be \( I = \int_{0}^{a} (a^2-x^2)^{3/2} dx \).
We use a trigonometric substitution. Let \( x = a \sin \theta \).
Then \( dx = a \cos \theta \, d\theta \).
When \( x = 0 \), \( 0 = a \sin \theta \implies \theta = 0 \).
When \( x = a \), \( a = a \sin \theta \implies \sin \theta = 1 \implies \theta = \frac{\pi}{2} \).
Substitute these into the integral:
\( I = \int_{0}^{\pi/2} (\sqrt{a^2-(a \sin \theta)^2})^3 (a \cos \theta) d\theta \)
\( = \int_{0}^{\pi/2} (\sqrt{a^2-a^2 \sin^2 \theta})^3 (a \cos \theta) d\theta \)
\( = \int_{0}^{\pi/2} (\sqrt{a^2(1-\sin^2 \theta)})^3 (a \cos \theta) d\theta \)
\( = \int_{0}^{\pi/2} (\sqrt{a^2 \cos^2 \theta})^3 (a \cos \theta) d\theta \)
\( = \int_{0}^{\pi/2} (a \cos \theta)^3 (a \cos \theta) d\theta \)
\( = \int_{0}^{\pi/2} a^3 \cos^3 \theta \cdot a \cos \theta \, d\theta \)
\( = a^4 \int_{0}^{\pi/2} \cos^4 \theta \, d\theta \)
Now, we use Wallis's Formula for \( \int_{0}^{\pi/2} \cos^n x \, dx \). For \( n=4 \), it is:
\( \int_{0}^{\pi/2} \cos^4 \theta \, d\theta = \frac{(4-1)(4-3)}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3}{8} \cdot \frac{\pi}{2} = \frac{3\pi}{16} \)
So, \( I = a^4 \cdot \frac{3\pi}{16} = \frac{3\pi a^4}{16} \)
In simple words: We solved this integral by changing the variable from 'x' to 'theta' using a trigonometric substitution. After simplifying, we used a special formula called Wallis's Formula to quickly solve the definite integral. This method helps when integrals involve square roots of squared terms.

🎯 Exam Tip: For integrals involving \( \sqrt{a^2-x^2} \), always think of trigonometric substitutions like \( x = a \sin \theta \) or \( x = a \cos \theta \) to simplify the expression and then apply Wallis's formula if the limits are \( 0 \) to \( \pi/2 \).

Question 20. If \( \int_{0}^{x} f(t) dt = x + \int_{x}^{1} f(t) dt \), then the value of f(1) is
(a) \( \frac { 1 }{ 2 } \)
(b) \( \frac { 1 }{ 3 } \)
(c) \( 1 \)
(d) \( \frac { 3 }{ 4 } \)
Answer: (a) \( \frac { 1 }{ 2 } \)
Hint:
We are given the equation: \( \int_{0}^{x} f(t) dt = x + \int_{x}^{1} f(t) dt \).
To find \( f(x) \), we need to differentiate both sides of the equation with respect to \( x \).
First, recall the Fundamental Theorem of Calculus, which states that \( \frac{d}{dx} \int_{a}^{x} g(t) dt = g(x) \).
Also, for \( \int_{x}^{a} g(t) dt \), we can write it as \( - \int_{a}^{x} g(t) dt \).
So, \( \int_{x}^{1} f(t) dt = - \int_{1}^{x} f(t) dt \).
Our equation becomes: \( \int_{0}^{x} f(t) dt = x - \int_{1}^{x} f(t) dt \).
Now, differentiate both sides with respect to \( x \):
\( \frac{d}{dx} \left( \int_{0}^{x} f(t) dt \right) = \frac{d}{dx} (x) - \frac{d}{dx} \left( \int_{1}^{x} f(t) dt \right) \)
Using the Fundamental Theorem of Calculus:
\( f(x) = 1 - f(x) \)
Now, we solve for \( f(x) \):
\( f(x) + f(x) = 1 \)
\( 2f(x) = 1 \)
\( f(x) = \frac{1}{2} \)
The question asks for the value of \( f(1) \). Since \( f(x) \) is a constant function \( \frac{1}{2} \), its value is the same for all \( x \).
Therefore, \( f(1) = \frac{1}{2} \).
In simple words: We took the given equation and used a rule called the Fundamental Theorem of Calculus. This rule helps us find the original function when we differentiate an integral. By doing this on both sides, we found that \( f(x) \) is always \( \frac{1}{2} \), so \( f(1) \) is also \( \frac{1}{2} \).

🎯 Exam Tip: When an integral equation involves limits that are variables, differentiating with respect to the variable limit (using the Fundamental Theorem of Calculus) is often the key to solving the problem. Pay close attention to the order of limits in the integral.