Samacheer Kalvi Class 12 Business Maths Solutions Chapter 2 Integral Calculus I Exercise 2.12

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 2 Integral Calculus I Ex 2.12

Choose the most suitable answer from the given four alternatives:

Question 1. \( \int\frac { 1 }{x^3} dx \) is
(a) \( \frac { -3 }{x^2} + c \)
(b) \( \frac {-1}{2x^2} + c \)
(c) \( \frac {-1}{3x^2} + c \)
(d) \( \frac { -1 }{x^2} + c \)
Answer: (b) \( \frac {-1}{2x^2} + c \)
\( \int\frac { 1 }{x^3} dx = \int x^{-3} dx \)
\( \implies = \frac {x^{-3+1} }{-3+1} + c \)
\( \implies = \frac {x^{-2} }{-2} + c \)
\( \implies = \frac { -1 }{2x^2} + c \) This shows how to integrate negative powers of x.
In simple words: To solve this, you add 1 to the power of x and then divide by the new power. A constant 'c' is added for indefinite integrals.

🎯 Exam Tip: Remember that \( \frac{1}{x^n} = x^{-n} \). This transformation is crucial for integrating rational functions.

Question 2. \( \int 2^x dx \) is
(a) \( 2^x \log 2 + c \)
(b) \( 2^x + c \)
(c) \( \frac { 2^x }{\log 2} + c \)
(d) \( \frac { \log 2 }{2^x} + c \)
Answer: (c) \( \frac { 2^x }{\log 2} + c \)
We use the standard integration formula for exponential functions.
The formula is \( \int a^x dx = \frac { a^x }{\log a} + c \). Here, 'a' is 2.
In simple words: When you integrate a number raised to the power of x, you get the same number to the power of x, divided by the natural logarithm of that number, plus a constant.

🎯 Exam Tip: Do not confuse the integration formula \( \int a^x dx = \frac { a^x }{\log a} \) with the differentiation formula \( \frac{d}{dx}(a^x) = a^x \log a \).

Question 3. \( \int \frac { \sin 2x }{2 \sin x} dx \) is
(a) \( \sin x + c \)
(b) \( \frac { 1 }{2} \sin x + c \)
(c) \( \cos x + c \)
(d) \( \frac { 1 }{2} \cos x + c \)
Answer: (a) \( \sin x + c \)
First, we use the trigonometric identity \( \sin 2x = 2 \sin x \cos x \). This helps simplify the expression.
\( \int \frac { \sin 2x }{2 \sin x} dx = \int \frac { 2 \sin x \cos x }{2 \sin x} dx \)
\( \implies = \int \cos x dx \)
\( \implies = \sin x + c \)
In simple words: We changed \( \sin 2x \) to \( 2 \sin x \cos x \), which let us cancel out \( 2 \sin x \), leaving only \( \cos x \) to integrate. The integral of \( \cos x \) is \( \sin x \).

🎯 Exam Tip: Always look for trigonometric identities to simplify expressions before integrating. This can turn a complex integral into a simple one.

Question 4. \( \int\frac { \sin 5x-\sin x }{\cos 3x} dx \) is
(a) \( -\cos 2x + c \)
(b) \( -\cos 2x - c \)
(c) \( -\frac { 1 }{4} \cos 2x + c \)
(d) \( -4 \cos 2x + c \)
Answer: (a) \( -\cos 2x + c \)
We use the sum-to-product trigonometric identity: \( \sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right) \).
Here, \( A = 5x \) and \( B = x \). So \( \frac{A+B}{2} = \frac{5x+x}{2} = \frac{6x}{2} = 3x \) and \( \frac{A-B}{2} = \frac{5x-x}{2} = \frac{4x}{2} = 2x \).
So, \( \sin 5x - \sin x = 2 \cos 3x \sin 2x \).
Now, substitute this into the integral:
\( \int \frac { 2 \cos 3x \sin 2x }{\cos 3x} dx \)
\( \implies = \int 2 \sin 2x dx \)
\( \implies = 2 \left( \frac{-\cos 2x}{2} \right) + c \)
\( \implies = -\cos 2x + c \)
In simple words: We used a special formula to change the top part of the fraction. After simplifying, we were left with \( 2 \sin 2x \), which we then integrated to get \( -\cos 2x \).

🎯 Exam Tip: Mastering trigonometric identities, especially sum-to-product and product-to-sum, is vital for simplifying complex integrals in trigonometry.

Question 5. \( \int \frac { \log x}{x} dx, x > 0 \) is
(a) \( \frac { 1 }{2} (\log x)^2 + c \)
(b) \( -\frac { 1 }{2} (\log x)^2 + c \)
(c) \( \frac { 2 }{x^2} + c \)
(d) \( \frac { 2 }{x^2} - c \)
Answer: (a) \( \frac { 1 }{2} (\log x)^2 + c \)
We use the substitution method to solve this integral.
Let \( t = \log x \).
Then, differentiate t with respect to x: \( \frac { dt }{dx} = \frac { 1 }{x} \).
This means \( dt = \frac { 1 }{x} dx \).
Now substitute 't' and 'dt' into the integral:
\( \int t dt \)
\( \implies = \frac { t^2 }{2} + c \)
Finally, substitute \( t = \log x \) back into the result:
\( \implies = \frac { (\log x)^2 }{2} + c \)
In simple words: We replaced \( \log x \) with 't' and \( \frac{1}{x} dx \) with 'dt'. Then we integrated 't' which gave \( t^2/2 \), and finally put \( \log x \) back in place of 't'.

🎯 Exam Tip: The key to solving integrals of the form \( \int \frac{f'(x)}{f(x)} dx \) or \( \int f(x) f'(x) dx \) is often to use the substitution method by setting \( t = f(x) \).

Question 6. \( \int\frac { e^x }{\sqrt{1+e^x}} dx \) is
(a) \( \frac { e^x }{\sqrt{1+e^x}} + c \)
(b) \( 2\sqrt{1+e^x} + c \)
(c) \( \sqrt{1+e^x} + c \)
(d) \( e^x\sqrt{1+e^x} + c \)
Answer: (b) \( 2\sqrt{1+e^x} + c \)
We use the substitution method.
Let \( t = 1+e^x \).
Then, \( \frac { dt }{dx} = e^x \).
This gives us \( dt = e^x dx \).
Now, substitute 't' and 'dt' into the integral:
\( \int \frac { dt }{\sqrt{t}} \)
\( \implies = \int t^{-1/2} dt \)
\( \implies = \frac { t^{-1/2+1} }{ -1/2+1 } + c \)
\( \implies = \frac { t^{1/2} }{ 1/2 } + c \)
\( \implies = 2t^{1/2} + c \)
\( \implies = 2\sqrt{t} + c \)
Finally, substitute \( t = 1+e^x \) back into the result:
\( \implies = 2\sqrt{1+e^x} + c \)
In simple words: We replaced the part under the square root with 't'. This made the integral simpler, turning it into \( \int t^{-1/2} dt \). After integrating and putting 't' back, we got the answer.

🎯 Exam Tip: When you see an expression and its derivative in an integral, consider substitution. For example, if you see \( e^x \) and \( e^x \) again (or its derivative) in the numerator, substitution with \( 1+e^x \) is a good approach.

Question 7. \( \int\sqrt { e^x} dx \) is
(a) \( \sqrt { e^x} + c \)
(b) \( 2\sqrt { e^x} + c \)
(c) \( \frac { 1 }{2} \sqrt { e^x} + c \)
(d) \( \frac { 1 }{2\sqrt { e^x}} + c \)
Answer: (b) \( 2\sqrt { e^x} + c \)
First, rewrite the square root of \( e^x \) as a power.
\( \sqrt { e^x} = (e^x)^{1/2} = e^{x/2} \).
Now integrate \( e^{x/2} \):
\( \int e^{x/2} dx \)
\( \implies = \frac { e^{x/2} }{1/2} + c \)
\( \implies = 2e^{x/2} + c \)
\( \implies = 2(e^x)^{1/2} + c \)
\( \implies = 2\sqrt { e^x} + c \)
In simple words: We changed the square root of \( e^x \) into \( e \) to the power of \( x/2 \). Then we integrated it by dividing by the power of x, which is \( 1/2 \), and got \( 2\sqrt{e^x} \).

🎯 Exam Tip: Remember that \( \sqrt{f(x)} = (f(x))^{1/2} \). This power form helps in applying standard integration rules, especially for exponential functions.

Question 8. \( \int e^{2x} [2x^2 + 2x] dx \) is
(a) \( e^{2x} x^2 + c \)
(b) \( xe^{2x} + c \)
(c) \( 2x^2e^2 + c \)
(d) \( \frac {x^2e^x }{2} + c \)
Answer: (a) \( e^{2x} x^2 + c \)
This integral is of the form \( \int e^{ax} [a f(x) + f'(x)] dx \), which integrates to \( e^{ax} f(x) + c \).
Here, compare \( \int e^{2x} [2x^2 + 2x] dx \) with the general form.
Let \( a = 2 \).
Let \( f(x) = x^2 \).
Then, \( f'(x) = 2x \).
So the expression \( [2x^2 + 2x] \) can be written as \( [a f(x) + f'(x)] \).
Therefore, the integral is \( e^{2x} x^2 + c \). This property simplifies integrals involving products of exponential and other functions.
In simple words: This integral has a special pattern where we can see a function \( x^2 \) and its derivative \( 2x \), both multiplied by \( e^{2x} \). So, the answer is just \( e^{2x} \) multiplied by the original function \( x^2 \).

🎯 Exam Tip: Recognize the integral form \( \int e^{ax} [a f(x) + f'(x)] dx = e^{ax} f(x) + c \). This property is a powerful shortcut for specific types of integrals involving exponential functions and derivatives.

Question 9. \( \int \frac { e^x }{e^x+1} dx \) is
(a) \( \log |\frac { e^x }{e^x+1}| + c \)
(b) \( \log |\frac { e^x+1 }{e^x}| + c \)
(c) \( \log |e^x| + c \)
(d) \( \log |e^x + 1| + c \)
Answer: (d) \( \log |e^x + 1| + c \)
We use the substitution method.
Let \( t = e^x + 1 \).
Then, differentiate t with respect to x: \( \frac { dt }{dx} = e^x \).
This means \( dt = e^x dx \).
Now, substitute 't' and 'dt' into the integral:
\( \int \frac { dt }{t} \)
\( \implies = \log |t| + c \)
Finally, substitute \( t = e^x + 1 \) back into the result:
\( \implies = \log |e^x + 1| + c \)
In simple words: We replaced the bottom part of the fraction with 't'. Since the top part was its derivative, the integral became \( \int \frac{1}{t} dt \), which is \( \log|t| \). Then we put back the original expression for 't'.

🎯 Exam Tip: When the numerator is the derivative of the denominator, the integral is always the natural logarithm of the absolute value of the denominator: \( \int \frac{f'(x)}{f(x)} dx = \log |f(x)| + c \).

Question 10. \( \int (\frac { 9 }{x-3}-\frac { 1 }{x+1}) dx \) is
(a) \( \log |x - 3| - \log|x + 1| + c \)
(b) \( \log|x - 3| + \log|x + 1| + c \)
(c) \( 9 \log |x - 3| - \log |x + 1| + c \)
(d) \( 9 \log |x - 3 + \log x + 1| + c \)
Answer: (c) \( 9 \log |x - 3| - \log |x + 1| + c \)
We can integrate each term separately using the formula \( \int \frac{1}{ax+b} dx = \frac{1}{a} \log|ax+b| \).
\( \int \frac { 9 }{x-3} dx - \int \frac { 1 }{x+1} dx \)
\( \implies = 9 \int \frac { 1 }{x-3} dx - 1 \int \frac { 1 }{x+1} dx \)
\( \implies = 9 \log |x-3| - \log |x+1| + c \)
This is a direct application of the logarithmic integration rule.
In simple words: We took the integral of each part separately. The integral of \( \frac{1}{x-3} \) is \( \log|x-3| \), and the integral of \( \frac{1}{x+1} \) is \( \log|x+1| \). The constant 9 just stays in front of its logarithm.

🎯 Exam Tip: Remember that integrals are linear. This means you can integrate each term of a sum or difference separately and factor out constants.

Question 11. \( \int \frac { 2x^3 }{4+x^4} dx \) is
(a) \( \log |4 + x^4] + c \)
(b) \( \frac { 1 }{2} \log |4 + x^4| + c \)
(c) \( \frac { 1 }{2} \log |4 + x^4 + c \)
(d) \( \log |\frac { 2x^3 }{4+x^4} + c \)
Answer: (b) \( \frac { 1 }{2} \log |4 + x^4| + c \)
We use the substitution method.
Let \( t = 4+x^4 \).
Then, \( \frac { dt }{dx} = 4x^3 \).
This implies \( dt = 4x^3 dx \).
We have \( 2x^3 dx \) in the numerator, so we can write \( 2x^3 dx = \frac{1}{2} dt \).
Now, substitute 't' and 'dt' into the integral:
\( \int \frac { \frac{1}{2} dt }{t} \)
\( \implies = \frac { 1 }{2} \int \frac { dt }{t} \)
\( \implies = \frac { 1 }{2} \log |t| + c \)
Finally, substitute \( t = 4+x^4 \) back into the result:
\( \implies = \frac { 1 }{2} \log |4 + x^4| + c \)
In simple words: We replaced the denominator with 't'. The numerator became \( \frac{1}{2} dt \). So, we integrated \( \frac{1}{2} \times \frac{1}{t} \), which gave \( \frac{1}{2} \log|t| \). Then we put back the original expression for 't'.

🎯 Exam Tip: Always check if the numerator is a multiple of the derivative of the denominator. If so, a simple substitution \( t = \text{denominator} \) will simplify the integral to a logarithmic form.

Question 12. \( \int \frac { dx }{\sqrt{x^2-36}} \) is
(a) \( \sqrt{x^2-36} + c \)
(b) \( \log |x + \sqrt{x^2-36}| + c \)
(c) \( \log |x-\sqrt{x^2-36}| + c \)
(d) \( \log |x^2 + \sqrt{x^2-36}| + c \)
Answer: (b) \( \log |x + \sqrt{x^2-36}| + c \)
This is a standard integral form. We compare it to the general formula \( \int \frac{dx}{\sqrt{x^2-a^2}} = \log|x + \sqrt{x^2-a^2}| + c \).
In this problem, \( a^2 = 36 \), so \( a = 6 \).
Therefore, by direct application of the formula:
\( \int \frac { dx }{\sqrt{x^2-36}} = \int \frac { dx }{\sqrt{x^2-6^2}} \)
\( \implies = \log |x + \sqrt{x^2-6^2}| + c \)
\( \implies = \log |x + \sqrt{x^2-36}| + c \)
Recognizing standard integral forms helps solve problems quickly.
In simple words: This integral matches a known formula for \( \frac{1}{\sqrt{x^2-a^2}} \). We just need to find the value of 'a' and put it into the formula, which gives us a logarithm.

🎯 Exam Tip: Memorize the standard integral formulas involving square roots, such as \( \int \frac{dx}{\sqrt{x^2-a^2}} \), \( \int \frac{dx}{\sqrt{a^2-x^2}} \), and \( \int \frac{dx}{\sqrt{x^2+a^2}} \), as they are frequently tested.

Question 13. \( \int\frac { 2x+3 }{\sqrt{x^2+3x+2}} dx \) is
(a) \( \sqrt{x^2+3x+2} + c \)
(b) \( 2\sqrt{x^2+3x+2} + c \)
(c) \( \sqrt{x^2+3x+2} + c \)
(d) \( \frac { 2 }{3} (x^2 + 3x + 2) + c \)
Answer: (b) \( 2\sqrt{x^2+3x+2} + c \)
We use the substitution method.
Let \( t = x^2+3x+2 \).
Then, differentiate t with respect to x: \( \frac { dt }{dx} = 2x+3 \).
This means \( dt = (2x+3) dx \).
Now, substitute 't' and 'dt' into the integral:
\( \int \frac { dt }{\sqrt{t}} \)
\( \implies = \int t^{-1/2} dt \)
\( \implies = \frac { t^{-1/2+1} }{ -1/2+1 } + c \)
\( \implies = \frac { t^{1/2} }{ 1/2 } + c \)
\( \implies = 2t^{1/2} + c \)
\( \implies = 2\sqrt{t} + c \)
Finally, substitute \( t = x^2+3x+2 \) back into the result:
\( \implies = 2\sqrt{x^2+3x+2} + c \)
In simple words: We replaced the expression inside the square root with 't'. The numerator became 'dt'. This simplified the integral to \( \int t^{-1/2} dt \), which, after integrating, became \( 2\sqrt{t} \). We then put back the original expression for 't'.

🎯 Exam Tip: Always check if the numerator is the derivative of the expression inside a square root or power in the denominator. This suggests a direct substitution method.

Question 14. \( \int_{0}^{4} (2x + 1) dx \) is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (b) 2
First, find the indefinite integral of \( (2x+1) \).
\( \int (2x+1) dx = 2 \frac{x^2}{2} + x + C = x^2 + x + C \).
Now, evaluate the definite integral using the limits from 0 to 1 (as per the source's calculation leading to the given answer, despite the question stating 0 to 4):
\( [x^2 + x]_{0}^{1} \)
\( \implies = [(1)^2 + (1)] - [(0)^2 + (0)] \)
\( \implies = [1 + 1] - [0] \)
\( \implies = 2 \)
The definite integral represents the area under the curve between the given limits. If the question intended bounds 0 to 4, the answer would be 20, which is not an option.
In simple words: We found the integral of \( 2x+1 \), which is \( x^2+x \). Then we put the upper limit (1) into this expression and subtracted what we got by putting the lower limit (0). This gave us 2.

🎯 Exam Tip: For definite integrals, carefully apply the Fundamental Theorem of Calculus: \( \int_a^b f(x) dx = F(b) - F(a) \), where \( F(x) \) is the antiderivative of \( f(x) \).

Question 15. \( \int_{2}^{4} \frac { dx }{x} \) is
(a) \( \log 4 \)
(b) 0
(c) \( \log 2 \)
(d) \( \log 8 \)
Answer: (c) \( \log 2 \)
First, find the indefinite integral of \( \frac{1}{x} \).
\( \int \frac { dx }{x} = \log |x| + C \).
Now, evaluate the definite integral using the limits from 2 to 4:
\( [\log |x|]_{2}^{4} \)
\( \implies = \log |4| - \log |2| \)
Using the logarithm property \( \log A - \log B = \log (\frac{A}{B}) \):
\( \implies = \log \left( \frac{4}{2} \right) \)
\( \implies = \log 2 \)
In simple words: We know the integral of \( \frac{1}{x} \) is \( \log|x| \). So, we just put in the top number (4) and the bottom number (2) into \( \log|x| \) and subtracted the results.

🎯 Exam Tip: Remember the basic integral \( \int \frac{1}{x} dx = \log|x| + C \) and the logarithm property \( \log A - \log B = \log(A/B) \).

Question 16. \( \int_{0}^{\infty} e^{-2x} dx \) is
(a) 0
(b) 1
(c) 2
(d) \( \frac { 1 }{2} \)
Answer: (d) \( \frac { 1 }{2} \)
First, find the indefinite integral of \( e^{-2x} \).
\( \int e^{-2x} dx = \frac { e^{-2x} }{ -2 } + C \).
Now, evaluate the definite integral using the limits from 0 to \( \infty \). Remember that \( e^{-\infty} = 0 \).
\( \left[ \frac { e^{-2x} }{ -2 } \right]_{0}^{\infty} \)
\( \implies = -\frac { 1 }{2} [e^{-2x}]_{0}^{\infty} \)
\( \implies = -\frac { 1 }{2} [e^{-2(\infty)} - e^{-2(0)}] \)
\( \implies = -\frac { 1 }{2} [0 - e^0] \)
\( \implies = -\frac { 1 }{2} [0 - 1] \)
\( \implies = -\frac { 1 }{2} (-1) \)
\( \implies = \frac { 1 }{2} \)
This type of integral is called an improper integral, often used in probability and physics.
In simple words: We integrated \( e^{-2x} \) and then put the limits \( \infty \) and 0. Because \( e \) to a very large negative power is zero, and \( e \) to the power of 0 is 1, we got \( \frac{1}{2} \).

🎯 Exam Tip: When evaluating improper integrals with an upper limit of infinity, remember that \( \lim_{x \to \infty} e^{-kx} = 0 \) for \( k > 0 \).

Question 17. \( \int_{-1}^{1} x^3 e^{x^4} dx \) is
(a) 1
(b) \( 2\int_{0}^{1} x^3 e^{x^4} \)
(c) 0
(d) \( e^4 \)
Answer: (c) 0
We can determine this by checking if the function is odd or even.
Let \( f(x) = x^3 e^{x^4} \).
Now, find \( f(-x) \):
\( f(-x) = (-x)^3 e^{(-x)^4} \)
\( \implies = -x^3 e^{x^4} \)
Since \( f(-x) = -f(x) \), the function \( f(x) \) is an odd function.
For an odd function, the definite integral over a symmetric interval \( [-a, a] \) is always zero.
\( \int_{-a}^{a} f(x) dx = 0 \) if \( f(x) \) is odd.
Thus, \( \int_{-1}^{1} x^3 e^{x^4} dx = 0 \). This property greatly simplifies certain definite integrals.
In simple words: We checked if the function was "odd" or "even". Since putting \( -x \) instead of \( x \) just changes the sign of the whole function, it's an odd function. For odd functions integrated from \( -1 \) to \( 1 \), the answer is always zero.

🎯 Exam Tip: Always check for symmetry in the integration limits. If the limits are \( -a \) to \( a \), determine if the function is odd or even to quickly find if the integral is 0 or \( 2 \times \int_0^a f(x) dx \).

Question 18. If \( f(x) \) is a continuous function and \( a < c < b \), then \( \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx \) is
(a) \( \int_{a}^{b} f(x) dx - \int_{a}^{c} f(x) dx \)
(b) \( \int_{a}^{c} f(x) dx - \int_{a}^{b} f(x) dx \)
(c) \( \int_{a}^{b} f(x) dx \)
(d) 0
Answer: (c) \( \int_{a}^{b} f(x) dx \)
This is a fundamental property of definite integrals. If a function \( f(x) \) is continuous on the interval \( [a, b] \) and \( c \) is any point in \( [a, b] \), then:
\( \int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx \).
This property is often called the additivity property of definite integrals with respect to the interval of integration.
In simple words: If you add the integral of a function from 'a' to 'c' and the integral from 'c' to 'b', it's the same as finding the integral directly from 'a' to 'b'.

🎯 Exam Tip: Understand the properties of definite integrals, especially the interval additivity property. It's crucial for solving problems where the integration range is split.

Question 19. The value of \( \int_{-\pi/2}^{\pi/2} \cos x dx \) is
(a) 0
(b) 2
(c) 1
(d) 4
Answer: (b) 2
We first check if \( f(x) = \cos x \) is an even or odd function.
\( f(-x) = \cos (-x) \). Since \( \cos(-x) = \cos x \), we have \( f(-x) = f(x) \).
Therefore, \( f(x) = \cos x \) is an even function.
For an even function integrated over a symmetric interval \( [-a, a] \), the property is:
\( \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx \).
So, \( \int_{-\pi/2}^{\pi/2} \cos x dx = 2 \int_{0}^{\pi/2} \cos x dx \).
Now, find the integral of \( \cos x \): \( \int \cos x dx = \sin x + C \).
Evaluate the definite integral:
\( = 2 [\sin x]_{0}^{\pi/2} \)
\( \implies = 2 [\sin (\pi/2) - \sin (0)] \)
\( \implies = 2 [1 - 0] \)
\( \implies = 2 \)
This property helps simplify calculations by changing the integration limits.
In simple words: \( \cos x \) is an "even" function, meaning it's symmetric. So, integrating it from \( -\pi/2 \) to \( \pi/2 \) is the same as integrating it from 0 to \( \pi/2 \) and then doubling the answer. We found the integral to be \( \sin x \), and when we put the limits, we got 2.

🎯 Exam Tip: For trigonometric functions over symmetric intervals, determine if they are odd or even. \( \cos x \) is even, \( \sin x \) is odd. This shortcut can save time in calculations.

Question 20. \( \int_{-\pi/2}^{\pi/2} \sqrt {x^4(1-x)^2} dx \)
(a) \( \frac { 1 }{12} \)
(b) \( \frac { -7 }{12} \)
(c) \( \frac { 1 }{20} \)
(d) \( \frac {-1}{12} \)
Answer: (a) \( \frac { 1 }{12} \)
We simplify the expression inside the square root:
\( \sqrt {x^4(1-x)^2} = \sqrt{(x^2)^2 (1-x)^2} = |x^2(1-x)| = x^2|1-x| \).
However, the calculation provided in the hint (and leading to the given answer) uses an integral from 0 to 1, effectively treating \( |1-x| \) as \( (1-x) \) for \( 0 \le x \le 1 \). Also, the limits in the question \( (-\pi/2, \pi/2) \) and the solution \( 1/12 \) are inconsistent. Following Iron Rule 6, we will reproduce the calculation from the hint that yields the given answer.
Let's assume the integral intended to be \( \int_{0}^{1} x^2(1-x) dx \):
\( \int_{0}^{1} (x^2 - x^3) dx \)
\( \implies = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{0}^{1} \)
\( \implies = \left( \frac{1^3}{3} - \frac{1^4}{4} \right) - \left( \frac{0^3}{3} - \frac{0^4}{4} \right) \)
\( \implies = \left( \frac{1}{3} - \frac{1}{4} \right) - 0 \)
\( \implies = \frac{4-3}{12} = \frac{1}{12} \)
In simple words: We simplified the square root part to \( x^2(1-x) \). Then we integrated \( x^2 - x^3 \) from 0 to 1. After putting in the numbers and subtracting, we got \( \frac{1}{12} \).

🎯 Exam Tip: When dealing with square roots of squared terms, remember to use absolute values, e.g., \( \sqrt{y^2} = |y| \). Also, be aware that definite integral problems may sometimes simplify to standard polynomial integrals over a specific range.

Question 21. If \( \int_{0}^{1} f(x) dx = 1 \), \( \int_{0}^{1} x f(x) dx = a \) and \( \int_{0}^{1} x^2 f(x) dx = a^2 \), then \( \int_{0}^{1} (a - x)^2 f(x) dx \) is
(a) \( 4a^2 \)
(b) 0
(c) \( a^2 \)
(d) 1
Answer: (b) 0
First, expand \( (a-x)^2 \):
\( (a-x)^2 = a^2 - 2ax + x^2 \).
Now, substitute this into the integral:
\( \int_{0}^{1} (a^2 - 2ax + x^2) f(x) dx \)
Using the linearity property of integrals, we can split this into three separate integrals:
\( \implies = \int_{0}^{1} a^2 f(x) dx - \int_{0}^{1} 2ax f(x) dx + \int_{0}^{1} x^2 f(x) dx \)
We can pull out constants from the integrals:
\( \implies = a^2 \int_{0}^{1} f(x) dx - 2a \int_{0}^{1} x f(x) dx + \int_{0}^{1} x^2 f(x) dx \)
Now, substitute the given values:
\( \implies = a^2(1) - 2a(a) + a^2 \)
\( \implies = a^2 - 2a^2 + a^2 \)
\( \implies = 0 \)
This problem demonstrates how to combine known integral values using properties of integrals.
In simple words: We expanded \( (a-x)^2 \) and then used the given integral values for \( f(x) \), \( xf(x) \), and \( x^2f(x) \). After putting in these values and doing the math, everything canceled out to zero.

🎯 Exam Tip: Remember that integrals are linear: \( \int [k_1 f(x) \pm k_2 g(x)] dx = k_1 \int f(x) dx \pm k_2 \int g(x) dx \). This allows you to break down complex integrals into simpler parts.

Question 22. The value of \( \int_{2}^{3} f(5 - x) dx - \int_{2}^{3} f(x) dx \) is
(a) 1
(b) 0
(c) -1
(d) 5
Answer: (b) 0
We use the property of definite integrals: \( \int_{a}^{b} f(a+b-x) dx = \int_{a}^{b} f(x) dx \).
In this problem, the limits are \( a=2 \) and \( b=3 \). So, \( a+b-x = 2+3-x = 5-x \).
Therefore, \( \int_{2}^{3} f(5-x) dx = \int_{2}^{3} f(x) dx \).
Now, substitute this back into the given expression:
\( \int_{2}^{3} f(5-x) dx - \int_{2}^{3} f(x) dx \)
\( \implies = \int_{2}^{3} f(x) dx - \int_{2}^{3} f(x) dx \)
\( \implies = 0 \)
This property is very useful for simplifying definite integrals.
In simple words: There's a rule that says if you change \( x \) to \( a+b-x \) inside an integral, the answer stays the same. So, the first integral became equal to the second integral, and when we subtracted them, we got zero.

🎯 Exam Tip: The property \( \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx \) is powerful. It allows you to transform an integral into a potentially easier form without changing its value.

Question 23. \( \int_{0}^{4} (\sqrt{x} + \frac { 1 }{\sqrt{x}}) dx \) is
(a) \( \frac { 20 }{3} \)
(b) \( \frac { 21 }{3} \)
(c) \( \frac { 28 }{3} \)
(d) \( \frac { 1 }{3} \)
Answer: (c) \( \frac { 28 }{3} \)
First, rewrite the terms with fractional exponents:
\( \sqrt{x} = x^{1/2} \)
\( \frac{1}{\sqrt{x}} = x^{-1/2} \)
Now, integrate each term separately:
\( \int (x^{1/2} + x^{-1/2}) dx \)
\( \implies = \frac { x^{1/2+1} }{1/2+1} + \frac { x^{-1/2+1} }{-1/2+1} + C \)
\( \implies = \frac { x^{3/2} }{3/2} + \frac { x^{1/2} }{1/2} + C \)
\( \implies = \frac { 2 }{3} x^{3/2} + 2x^{1/2} + C \)
Now, evaluate the definite integral from 0 to 4:
\( \left[ \frac { 2 }{3} x^{3/2} + 2x^{1/2} \right]_{0}^{4} \)
\( \implies = \left( \frac { 2 }{3} (4)^{3/2} + 2(4)^{1/2} \right) - \left( \frac { 2 }{3} (0)^{3/2} + 2(0)^{1/2} \right) \)
\( \implies = \left( \frac { 2 }{3} (8) + 2(2) \right) - (0 + 0) \)
\( \implies = \left( \frac { 16 }{3} + 4 \right) \)
\( \implies = \frac { 16 }{3} + \frac { 12 }{3} \)
\( \implies = \frac { 28 }{3} \)
In simple words: We changed the square roots into powers of \( x \). Then we integrated both parts separately. After putting in the upper limit (4) and the lower limit (0) into the integrated expression and subtracting, we got \( \frac{28}{3} \).

🎯 Exam Tip: Always convert square roots and fractions involving powers of x into exponential form (e.g., \( \sqrt{x} = x^{1/2} \), \( \frac{1}{x^n} = x^{-n} \)) before applying the power rule of integration.

Question 24. \( \int_{0}^{\pi/3} \tan x dx \) is
(a) \( \log 2 \)
(b) 0
(c) \( \log \sqrt{2} \)
(d) \( 2 \log 2 \)
Answer: (a) \( \log 2 \)
First, find the indefinite integral of \( \tan x \).
We know that \( \tan x = \frac{\sin x}{\cos x} \).
Let \( t = \cos x \). Then \( dt = -\sin x dx \), so \( -dt = \sin x dx \).
\( \int \tan x dx = \int \frac{\sin x}{\cos x} dx = \int \frac{-dt}{t} = -\log|t| + C \)
\( \implies = -\log|\cos x| + C \)
This can also be written as \( \log|\sec x| + C \) (since \( -\log A = \log(1/A) \)).
Now, evaluate the definite integral using the limits from 0 to \( \pi/3 \):
\( [\log|\sec x|]_{0}^{\pi/3} \)
\( \implies = \log|\sec (\pi/3)| - \log|\sec (0)| \)
We know \( \sec(\pi/3) = \frac{1}{\cos(\pi/3)} = \frac{1}{1/2} = 2 \).
And \( \sec(0) = \frac{1}{\cos(0)} = \frac{1}{1} = 1 \).
\( \implies = \log|2| - \log|1| \)
\( \implies = \log 2 - 0 \)
\( \implies = \log 2 \)
In simple words: We first found the integral of \( \tan x \), which is \( \log|\sec x| \). Then we put in the top limit \( (\pi/3) \) and the bottom limit (0) into this expression. After subtracting the results, we got \( \log 2 \).

🎯 Exam Tip: The integral of \( \tan x \) can be expressed as either \( -\log|\cos x| \) or \( \log|\sec x| \). Use the form that simplifies calculations with given limits.

Question 25. Using the factorial representation of the gamma function, which of the following is the solution for the gamma function \( \Gamma(n) \) when \( n = 8 \)
(a) 5040
(b) 5400
(c) 4500
(d) 5540
Answer: (a) 5040
For a positive integer \( n \), the gamma function \( \Gamma(n) \) is defined as \( \Gamma(n) = (n-1)! \).
In this question, \( n = 8 \).
So, \( \Gamma(8) = (8-1)! = 7! \).
Now, calculate the factorial of 7:
\( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)
\( \implies = 5040 \)
The gamma function extends the concept of factorials to real and complex numbers.
In simple words: The gamma function for a whole number \( n \) is simply the factorial of \( n-1 \). So for \( n=8 \), we calculated \( 7! \), which is \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \), and that gave us 5040.

🎯 Exam Tip: Remember the basic definition of the Gamma function for positive integers: \( \Gamma(n) = (n-1)! \). This relationship is frequently used in problems involving the Gamma function.

Question 26. \( \Gamma(n) \) is
(a) \( (n - 1)! \)
(b) \( n! \)
(c) \( n \Gamma (n) \)
(d) \( (n - 1) \Gamma(n) \)
Answer: (a) \( (n - 1)! \)
For a positive integer \( n \), the gamma function \( \Gamma(n) \) is defined as \( \Gamma(n) = (n-1)! \).
This is the fundamental relationship between the gamma function and factorials.
The recursive property of the gamma function is \( \Gamma(n) = (n-1)\Gamma(n-1) \). For integer \( n \), repeated application of this property leads to \( (n-1)! \).
In simple words: For any positive whole number 'n', the gamma function \( \Gamma(n) \) is simply equal to the factorial of \( n-1 \).

🎯 Exam Tip: This is a definition. Be sure to know it precisely for positive integers, as it's a key concept in the study of the Gamma function.

Question 27. \( \Gamma(1) \) is
(a) 0
(b) 1
(c) n
(d) \( n! \)
Answer: (b) 1
Using the definition \( \Gamma(n) = (n-1)! \) for positive integers \( n \).
For \( n = 1 \):
\( \Gamma(1) = (1-1)! = 0! \).
By definition, \( 0! = 1 \).
Therefore, \( \Gamma(1) = 1 \). This value is important as it sets a base for calculating other gamma function values for integers.
In simple words: Using the rule that \( \Gamma(n) \) is \( (n-1)! \), for \( n=1 \), we get \( 0! \). The value of \( 0! \) is always 1.

🎯 Exam Tip: Remember that \( 0! = 1 \) by convention, which makes \( \Gamma(1) = 1 \). This is a crucial starting point for gamma function calculations.

Question 28. If \( n > 0 \), then \( \Gamma(n) \) is
(a) \( \int_{0}^{1} e^{-x} x^{n-1} dx \)
(b) \( \int_{0}^{1} e^{-x} x^n dx \)
(c) \( \int_{0}^{\infty} e^x x^n dx \)
(d) \( \int_{0}^{\infty} e^{-x} x^{n-1} dx \)
Answer: (d) \( \int_{0}^{\infty} e^{-x} x^{n-1} dx \)
This is the integral definition of the Gamma function, which is valid for \( n > 0 \).
The integral representation of the Gamma function is:
\( \Gamma(n) = \int_{0}^{\infty} t^{n-1} e^{-t} dt \).
Replacing the dummy variable 't' with 'x' gives the option (d). This definition extends the factorial concept to non-integer values.
In simple words: This is the exact definition of the Gamma function using an integral. It means that the Gamma function of 'n' is found by integrating \( x \) to the power of \( n-1 \) multiplied by \( e \) to the power of \( -x \), from 0 to infinity.

🎯 Exam Tip: Commit the integral definition of the Gamma function to memory. It's fundamental for understanding the function beyond integer values.

Question 29. \( \Gamma\left(\frac { 3 }{2}\right) \)
(a) \( \sqrt{\pi} \)
(b) \( \frac { \sqrt{\pi} }{2} \)
(c) \( 2\sqrt{\pi} \)
(d) \( \frac { 3 }{2} \)
Answer: (b) \( \frac { \sqrt{\pi} }{2} \)
We use the recursive property of the Gamma function: \( \Gamma(n) = (n-1)\Gamma(n-1) \).
Also, we know that \( \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi} \).
For \( n = \frac{3}{2} \):
\( \Gamma\left(\frac{3}{2}\right) = \left(\frac{3}{2}-1\right) \Gamma\left(\frac{3}{2}-1\right) \)
\( \implies = \left(\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right) \)
\( \implies = \frac{1}{2} \sqrt{\pi} \)
\( \implies = \frac{\sqrt{\pi}}{2} \)
This shows how the recursive property can be used to calculate Gamma function values for non-integers.
In simple words: We used a special rule that says \( \Gamma(n) \) is \( (n-1) \) multiplied by \( \Gamma(n-1) \). We also know that \( \Gamma(\frac{1}{2}) \) is \( \sqrt{\pi} \). Putting these together, we found that \( \Gamma(\frac{3}{2}) \) is \( \frac{\sqrt{\pi}}{2} \).

🎯 Exam Tip: Remember the recursive relation \( \Gamma(n) = (n-1)\Gamma(n-1) \) and the special value \( \Gamma(1/2) = \sqrt{\pi} \). These are essential for evaluating Gamma functions for half-integer values.

Question 30. \( \int_{0}^{\infty} x^4 e^{-x} dx \) is
(a) 12
(b) 4
(c) \( 4! \)
(d) 64
Answer: (c) \( 4! \)
This integral is in the form of the Gamma function definition, \( \Gamma(n) = \int_{0}^{\infty} x^{n-1} e^{-x} dx \).
Comparing \( \int_{0}^{\infty} x^4 e^{-x} dx \) with \( \int_{0}^{\infty} x^{n-1} e^{-x} dx \):
We have \( n-1 = 4 \), which means \( n = 5 \).
So, the integral is equal to \( \Gamma(5) \).
Using the property \( \Gamma(n) = (n-1)! \) for positive integers:
\( \Gamma(5) = (5-1)! = 4! \).
Now, calculate the factorial of 4:
\( 4! = 4 \times 3 \times 2 \times 1 = 24 \).
This is a direct application of the integral definition of the Gamma function.
In simple words: This integral is actually the definition of the Gamma function. For \( x^4 e^{-x} \), the 'n' in the Gamma function definition is 5. So the answer is \( \Gamma(5) \), which is the same as \( (5-1)! \) or \( 4! \).

🎯 Exam Tip: Recognize integrals of the form \( \int_0^\infty x^{n-1} e^{-x} dx \) as \( \Gamma(n) \). This directly simplifies to \( (n-1)! \) when n is a positive integer.

Question 25. Using the factorial representation of the gamma function, which of the following is the solution for the gamma function \( \Gamma(n) \) when \( n = 8 \)
(a) 5040
(b) 5400
(c) 4500
(d) 5540
Answer: (a) 5040
In simple words: The gamma function \( \Gamma(n) \) for a positive integer \( n \) is defined as \( (n-1)! \). So for \( n=8 \), it is \( 7! \), which means multiplying all whole numbers from 7 down to 1. This calculation gives 5040.

🎯 Exam Tip: Remember the basic definition for integer values of \( n \): \( \Gamma(n) = (n-1)! \). For example, \( \Gamma(1) = 0! = 1 \) and \( \Gamma(4) = 3! = 6 \).

Question 26. \( \Gamma(n) \) is
(a) \( (n - 1)! \)
(b) \( n! \)
(c) \( n \Gamma (n) \)
(d) \( (n - 1) \Gamma(n) \)
Answer: (a) (n - 1)!
In simple words: The gamma function \( \Gamma(n) \) for a positive integer \( n \) is equal to \( (n-1)! \). It is a way to extend the idea of factorials to non-integer and complex numbers.

🎯 Exam Tip: This is a fundamental definition of the gamma function for positive integers. Ensure you know it for quick recall in MCQ questions.

Question 27. \( \Gamma(1) \) is
(a) 0
(b) 1
(c) n
(d) \( n! \)
Answer: (b) 1
In simple words: Using the definition \( \Gamma(n) = (n-1)! \), when \( n=1 \), we get \( (1-1)! = 0! \). By definition, \( 0! \) is equal to 1.

🎯 Exam Tip: Always remember that \( 0! = 1 \). This is a common point of confusion but is crucial for correct calculations involving factorials and the gamma function.

Question 28. If \( n > 0 \), then \( \Gamma(n) \) is
(a) \( \int_{0}^{1} e^{-x} x^{n-1} dx \)
(b) \( \int_{0}^{1} e^{-x} x^{n} dx \)
(c) \( \int_{0}^{\infty} e^{x} x^{n} dx \)
(d) \( \int_{0}^{\infty} e^{-x} x^{n-1} dx \)
Answer: (d) \( \int_{0}^{\infty} e^{-x} x^{n-1} dx \)
In simple words: The gamma function \( \Gamma(n) \) is formally defined by an integral. For any number \( n \) greater than zero, it is the integral from zero to infinity of \( e^{-x} \) multiplied by \( x \) raised to the power of \( (n-1) \). This integral helps calculate gamma values for non-integer numbers.

🎯 Exam Tip: This is the integral definition of the gamma function. Pay close attention to the limits of integration (0 to infinity) and the exponent of \( x \) being \( n-1 \).

Question 29. \( \Gamma\left(\frac{3}{2}\right) \)
(a) \( \sqrt{\pi} \)
(b) \( \frac{\sqrt{\pi}}{2} \)
(c) \( 2\sqrt{\pi} \)
(d) \( \frac{3}{2} \)
Answer: (b) \( \frac{\sqrt{\pi}}{2} \)
In simple words: We know that \( \Gamma(n+1) = n\Gamma(n) \) and \( \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi} \). To find \( \Gamma\left(\frac{3}{2}\right) \), we can write it as \( \Gamma\left(\frac{1}{2} + 1\right) \). Using the property, this becomes \( \frac{1}{2}\Gamma\left(\frac{1}{2}\right) \), which simplifies to \( \frac{\sqrt{\pi}}{2} \).

🎯 Exam Tip: Remember the two key values: \( \Gamma(1) = 1 \) and \( \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi} \). Use the recurrence relation \( \Gamma(n+1) = n\Gamma(n) \) to find other fractional gamma values.

Question 30. \( \int_{0}^{\infty} x^4 e^{-x} dx \) is
(a) 12
(b) 4
(c) \( 4! \)
(d) 64
Answer: (c) \( 4! \)
In simple words: This integral is a special form of the gamma function, where \( \Gamma(n) = \int_{0}^{\infty} x^{n-1} e^{-x} dx \). In this question, \( n-1 = 4 \), so \( n = 5 \). This means the integral is equal to \( \Gamma(5) \). For a positive integer \( n \), \( \Gamma(n) = (n-1)! \). So, \( \Gamma(5) = (5-1)! = 4! \).

🎯 Exam Tip: Recognize the integral form of the gamma function. Here, \( n-1 \) in the general formula corresponds to the power of \( x \) in the given integral. So, if \( x^k \) is in the integral, the answer is \( \Gamma(k+1) \) or \( k! \).