CBSE Class 12 Mathematics Integrals Assignment Set 02

Question 1. Evaluate \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^3 x \, dx\)
Answer: Let \(I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^3 x \, dx\)
Let \(f(x) = \sin^3 x\)
Then \(f(-x) = \sin^3(-x) = (-\sin x)^3 = -\sin^3 x = -f(x)\)
Since \(f(x)\) is an odd function, by using the property \(\int_{-a}^{a} f(x) \, dx = 0\) if \(f(-x) = -f(x)\):
\(I = 0\)
In simple words: The integral of an odd function over symmetric limits around zero is always zero.

Exam Tip: Always check for odd or even functions using \(f(-x)\) when limits are in the form \([-a, a]\) to save time.

Question 2. Evaluate \(\int_{0}^{3} (|x - 1| + |x - 2| + |x|) \, dx\)
Answer: Let \(I = \int_{0}^{3} (|x - 1| + |x - 2| + |x|) \, dx\)
We break the interval \([0, 3]\) at the critical points where the absolute values change sign, which are \(x = 0, 1, 2, 3\):
\(I = \int_{0}^{1} (-(x - 1) - (x - 2) + x) \, dx + \int_{1}^{2} ((x - 1) - (x - 2) + x) \, dx + \int_{2}^{3} ((x - 1) + (x - 2) + x) \, dx\)
Simplify each integrand:
For \([0, 1]\): \(-x + 1 - x + 2 + x = -x + 3\)
For \([1, 2]\): \(x - 1 - x + 2 + x = x + 1\)
For \([2, 3]\): \(x - 1 + x - 2 + x = 3x - 3\)
Now integrate each part separately:
\(I = \int_{0}^{1} (-x + 3) \, dx + \int_{1}^{2} (x + 1) \, dx + \int_{2}^{3} (3x - 3) \, dx\)
\(= \left[ -\frac{x^2}{2} + 3x \right]_{0}^{1} + \left[ \frac{x^2}{2} + x \right]_{1}^{2} + \left[ \frac{3x^2}{2} - 3x \right]_{2}^{3}\)
\(= \left( -\frac{1}{2} + 3 \right) + \left( \left(\frac{4}{2} + 2\right) - \left(\frac{1}{2} + 1\right) \right) + \left( \left(\frac{27}{2} - 9\right) - \left(\frac{12}{2} - 6\right) \right)\)
\(= \frac{5}{2} + \left( 4 - \frac{3}{2} \right) + \left( \frac{9}{2} - 0 \right) = \frac{5}{2} + \frac{5}{2} + \frac{9}{2} = \frac{19}{2}\)
In simple words: Split the integral at points where the expression inside the modulus becomes zero, then remove the absolute value signs accordingly.

Exam Tip: Be very careful with signs when breaking modulus integrals into sub-intervals.

Question 3. Evaluate \(\int_{-3}^{3} |x + 1| \, dx\)
Answer: Let \(I = \int_{-3}^{3} |x + 1| \, dx\)
The critical point is \(x = -1\), which lies inside the interval \([-3, 3]\).
Split the integral at \(x = -1\):
\(I = \int_{-3}^{-1} -(x + 1) \, dx + \int_{-1}^{3} (x + 1) \, dx\)
Integrate both parts:
\(= \left[ -\frac{(x+1)^2}{2} \right]_{-3}^{-1} + \left[ \frac{(x+1)^2}{2} \right]_{-1}^{3}\)
\(= \left( 0 - \left(-\frac{(-3+1)^2}{2}\right) \right) + \left( \frac{(3+1)^2}{2} - 0 \right)\)
\(= \left( \frac{(-2)^2}{2} \right) + \left( \frac{4^2}{2} \right) = \frac{4}{2} + \frac{16}{2} = 2 + 8 = 10\)
In simple words: Divide the range into parts where \(x + 1\) is negative and where it is positive.

Exam Tip: Sketching or mentally visualizing the graph of \(y = |x + 1|\) helps confirm integration limits.

Question 4. Evaluate \(\int_{0}^{\frac{\pi}{4}} \sqrt{1 + \sin 2x} \, dx\)
Answer: Let \(I = \int_{0}^{\frac{\pi}{4}} \sqrt{1 + \sin 2x} \, dx\)
Use trigonometric identities: \(1 = \sin^2 x + \cos^2 x\) and \(\sin 2x = 2\sin x \cos x\).
\(1 + \sin 2x = \sin^2 x + \cos^2 x + 2\sin x \cos x = (\sin x + \cos x)^2\)
Substitute back into the integral:
\(I = \int_{0}^{\frac{\pi}{4}} \sqrt{(\sin x + \cos x)^2} \, dx\)
For \(x \in \left[0, \frac{\pi}{4}\right]\), \(\sin x + \cos x \ge 0\), so:
\(I = \int_{0}^{\frac{\pi}{4}} (\sin x + \cos x) \, dx\)
Evaluate the integral:
\(I = \left[ -\cos x + \sin x \right]_{0}^{\frac{\pi}{4}}\)
\(= \left( -\cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) \right) - (-\cos(0) + \sin(0))\)
\(= \left( -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - (-1 + 0) = 0 - (-1) = 1\)
In simple words: Convert the expression under the square root into a complete square and simplify.

Exam Tip: Remember that \(\sqrt{a^2} = |a|\); verify the sign of the trigonometric function in the given interval.

Question 5. Evaluate \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \, dx\)
Answer: Let \(I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \, dx\)
Note that \(\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2\).
Let \(t = \sin x - \cos x\). Then \(dt = (\cos x + \sin x) \, dx\).
Change the limits of integration:
When \(x = \frac{\pi}{6}\), \(t = \sin\left(\frac{\pi}{6}\right) - \cos\left(\frac{\pi}{6}\right) = \frac{1}{2} - \frac{\sqrt{3}}{2} = \frac{1-\sqrt{3}}{2}\)
When \(x = \frac{\pi}{3}\), \(t = \sin\left(\frac{\pi}{3}\right) - \cos\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3}-1}{2}\)
Alternatively, use the substitution \(t = \sin x - \cos x\) and note that \(t^2 = (\sin x - \cos x)^2 = 1 - \sin 2x \implies \sin 2x = 1 - t^2\):
\(I = \int_{-\frac{\sqrt{3}-1}{2}}^{\frac{\sqrt{3}-1}{2}} \frac{1}{\sqrt{1 - t^2}} \, dt\)
Since the integrand is an even function and limits are symmetric \([-a, a]\):
\(I = 2 \int_{0}^{\frac{\sqrt{3}-1}{2}} \frac{1}{\sqrt{1 - t^2}} \, dt\)
\(= 2 \left[ \sin^{-1}(t) \right]_{0}^{\frac{\sqrt{3}-1}{2}} = 2 \sin^{-1}\left( \frac{\sqrt{3}-1}{2} \right)\)
In simple words: Substitute \(t = \sin x - \cos x\) so the numerator becomes the differential \(dt\).

Exam Tip: Recognizing standard derivative forms in the numerator is key to solving substitution problems quickly.

Question 6. Evaluate \(\int_{0}^{\pi} \frac{x \tan x}{\sec x + \tan x} \, dx\)
Answer: Let \(I = \int_{0}^{\pi} \frac{x \tan x}{\sec x + \tan x} \, dx \quad \text{--- (1)}\)
Rationalize the denominator or multiply numerator and denominator by \(\sec x - \tan x\):
\(\frac{\tan x}{\sec x + \tan x} = \frac{\tan x(\sec x - \tan x)}{\sec^2 x - \tan^2 x} = \tan x \sec x - \tan^2 x = \tan x \sec x - (\sec^2 x - 1)\)
Thus, \(I = \int_{0}^{\pi} x (\tan x \sec x - \sec^2 x + 1) \, dx\)
Using the property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx\):
\(I = \int_{0}^{\pi} (\pi - x) [\tan(\pi - x)\sec(\pi - x) - \sec^2(\pi - x) + 1] \, dx\)
\(I = \int_{0}^{\pi} (\pi - x) [-\tan x(-\sec x) - \sec^2 x + 1] \, dx\)
\(I = \int_{0}^{\pi} (\pi - x) (\tan x \sec x - \sec^2 x + 1) \, dx \quad \text{--- (2)}\)
Adding (1) and (2):
\(2I = \int_{0}^{\pi} \pi (\tan x \sec x - \sec^2 x + 1) \, dx\)
\(2I = \pi \int_{0}^{\pi} (\tan x \sec x - \sec^2 x + 1) \, dx\)
\(2I = \pi \left[ \sec x - \tan x + x \right]_{0}^{\pi}\)
Evaluate at limits:\(\sec(\pi) = -1, \tan(\pi) = 0\) and \(\sec(0) = 1, \tan(0) = 0\):
\(2I = \pi [(-1 - 0 + \pi) - (1 - 0 + 0)]\)
\(2I = \pi [\pi - 2]\)
\(I = \frac{\pi(\pi - 2)}{2}\)
In simple words: Use the king property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx\) to eliminate the \(x\) factor in the numerator.

Exam Tip: Whenever \(x\) multiplies a trigonometric function from \(0\) to \(\pi\), applying the reflection property is the standard approach.

Question 7. Prove that: \(\int_{0}^{1} \tan^{-1}\left(\frac{1}{1 - x + x^2}\right) \, dx = \frac{\pi}{2} - \log 2\)
Answer: Let \(I = \int_{0}^{1} \tan^{-1}\left(\frac{1}{1 - x + x^2}\right) \, dx\)
Rewrite the argument inside \(\tan^{-1}\) as: \(\frac{1}{1 + x(x-1)}\)
Using the formula \(\tan^{-1}\left(\frac{a - b}{1 + ab}\right) = \tan^{-1} a - \tan^{-1} b\):
\(\frac{1}{1 + x(x-1)} = \frac{x - (x-1)}{1 + x(x-1)}\)
Thus, the integrand becomes:
\(\tan^{-1}(x) - \tan^{-1}(x-1)\)
Now, apply the property \(\int_{0}^{1} f(x) \, dx = \int_{0}^{1} f(1-x) \, dx\):
\(I = \int_{0}^{1} \left[ \tan^{-1}(1-x) - \tan^{-1}(1-x-1) \right] \, dx\)
\(I = \int_{0}^{1} \left[ \tan^{-1}(1-x) - \tan^{-1}(-x) \right] \, dx\)
Since \(\tan^{-1}(-x) = -\tan^{-1}(x)\):
\(I = \int_{0}^{1} \left[ \tan^{-1}(1-x) + \tan^{-1}(x) \right] \, dx\)
Adding the original integral and the transformed integral using \(\int_{0}^{1} f(x) \, dx = \int_{0}^{1} f(1-x) \, dx\), we can also evaluate directly by splitting or noting that \(\int_{0}^{1} \tan^{-1}(x) \, dx + \int_{0}^{1} \tan^{-1}(1-x) \, dx\):
Let us use integration by parts for \(\int_{0}^{1} \tan^{-1}(x) \, dx\):
\(= [x \tan^{-1} x]_{0}^{1} - \int_{0}^{1} \frac{x}{1+x^2} \, dx\)
\(= 1 \cdot \frac{\pi}{4} - 0 - \frac{1}{2} \left[ \log(1+x^2) \right]_{0}^{1}\)
\(= \frac{\pi}{4} - \frac{1}{2}(\log 2 - \log 1) = \frac{\pi}{4} - \frac{1}{2}\log 2\)
Similarly, for \(\int_{0}^{1} \tan^{-1}(x-1) \, dx\), let \(u = x-1\), \(du = dx\):
\(\int_{-1}^{0} \tan^{-1}(u) \, du = [u \tan^{-1} u]_{-1}^{0} - \int_{-1}^{0} \frac{u}{1+u^2} \, du\)
\(= (0 - (-1)\left(-\frac{\pi}{4}\right)) - \frac{1}{2}[\log(1+u^2)]_{-1}^{0} = -\frac{\pi}{4} - \frac{1}{2}(0 - \log 2) = -\frac{\pi}{4} + \frac{1}{2}\log 2\)
Subtracting the two parts: \(I = \left(\frac{\pi}{4} - \frac{1}{2}\log 2\right) - \left(-\frac{\pi}{4} + \frac{1}{2}\log 2\right) = \frac{\pi}{2} - \log 2\)
In simple words: Express the fraction inside the inverse tangent as a difference of two terms and integrate using parts.

Exam Tip: Splitting inverse trigonometric integrals into standard logarithmic forms using integration by parts is a standard board exam technique.

Question 8. Evaluate \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (|\sin x| - |\cos x|) \, dx\)
Answer: Let \(I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (|\sin x| - |\cos x|) \, dx\)
Both \(|\sin x|\) and \(|\cos x|\) are even functions on \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Using the property for even functions: \(\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx\):
\(I = 2 \int_{0}^{\frac{\pi}{2}} (|\sin x| - |\cos x|) \, dx\)
In the interval \(\left[0, \frac{\pi}{2}\right]\), both \(\sin x \ge 0\) and \(\cos x \ge 0\):
\(I = 2 \left( \int_{0}^{\frac{\pi}{2}} \sin x \, dx - \int_{0}^{\frac{\pi}{2}} \cos x \, dx \right)\)
\(= 2 \left( [-\cos x]_{0}^{\frac{\pi}{2}} - [\sin x]_{0}^{\frac{\pi}{2}} \right)\)
\(= 2 \left( (-\cos\left(\frac{\pi}{2}\right) - (-\cos(0))) - (\sin\left(\frac{\pi}{2}\right) - \sin(0)) \right)\)
\(= 2 \left( (0 - (-1)) - (1 - 0) \right) = 2(1 - 1) = 0\)
In simple words: Simplify using even function properties and evaluate the basic trigonometric integrals.

Exam Tip: Notice symmetry to simplify definite integrals over symmetric intervals.

Question 9. Evaluate \(\int_{0}^{\pi} \frac{x \tan x}{\sec x \csc x} \, dx\)
Answer: Let \(I = \int_{0}^{\pi} \frac{x \tan x}{\sec x \csc x} \, dx\)
Simplify the denominator: \(\sec x \csc x = \frac{1}{\cos x \sin x} = \frac{2}{\sin 2x}\)
Also, \(\frac{\tan x}{\sec x \csc x} = \frac{\sin x}{\cos x} \cdot \sin x \cos x = \sin^2 x\)
Thus, the integral becomes:
\(I = \int_{0}^{\pi} x \sin^2 x \, dx \quad \text{--- (1)}\)
Apply the property \(\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx\):
\(I = \int_{0}^{\pi} (\pi - x) \sin^2(\pi - x) \, dx\)
\(I = \int_{0}^{\pi} (\pi - x) \sin^2 x \, dx \quad \text{--- (2)}\)
Adding (1) and (2):
\(2I = \int_{0}^{\pi} (\pi) \sin^2 x \, dx\)
\(2I = \pi \int_{0}^{\pi} \sin^2 x \, dx\)
Since \(\sin^2 x\) is symmetric about \(\frac{\pi}{2}\), use the property \(\int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx\) if \(f(2a-x) = f(x)\):
\(2I = \pi \left( 2 \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx \right)\)
\(I = \pi \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx\)
Using the reduction formula (Wallis's formula):
\(\int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}\)
Therefore, \(I = \pi \left( \frac{\pi}{4} \right) = \frac{\pi^2}{4}\)
In simple words: Simplify the trigonometric expression first, then apply the property to eliminate \(x\).

Exam Tip: Simplifying complex trigonometric denominators before applying integration properties makes the problem much easier.

Question 10. Using limit of sum evaluate \(\int_{a}^{b} (x + e^x + \cos x) \, dx\)
Answer: By definition of definite integral as limit of a sum:
\(\int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} h \sum_{r=0}^{n-1} f(a + rh)\), where \(h = \frac{b - a}{n}\).
Here \(f(x) = x + e^x + \cos x\):
\(f(a + rh) = (a + rh) + e^{a+rh} + \cos(a + rh)\)
Summing from \(r = 0\) to \(n-1\):
1. For linear term: \(\sum_{r=0}^{n-1} (a + rh) = na + h \frac{(n-1)n}{2} = (b-a)a + \frac{(b-a)^2}{n^2} \frac{(n-1)n}{2}\)
Taking limit as \(n \to \infty\), this evaluates to \(a(b-a) + \frac{(b-a)^2}{2} = \frac{b^2 - a^2}{2}\).
2. For exponential term: \(\sum_{r=0}^{n-1} e^{a+rh} = e^a \sum_{r=0}^{n-1} (e^h)^r = e^a \frac{e^{nh} - 1}{e^h - 1} = e^a \frac{e^{b-a} - 1}{\frac{e^h - 1}{h} \cdot h} \cdot h\)
Since \(nh = b-a\) and as \(n \to \infty, h \to 0\), \(\lim_{h \to 0} \frac{e^h - 1}{h} = 1\):
Limit is \(e^a (e^{b-a} - 1) = e^b - e^a\).
3. For trigonometric term: \(\sum_{r=0}^{n-1} \cos(a + rh) = \frac{\sin\left(\frac{nh}{2}\right)}{\sin\left(\frac{h}{2}\right)} \cos\left(a + \frac{(n-1)h}{2}\right)\)
Taking limit as \(n \to \infty\), this evaluates to \(\sin b - \sin a\).
Combining all parts:
\(\int_{a}^{b} (x + e^x + \cos x) \, dx = \left[ \frac{x^2}{2} + e^x + \sin x \right]_{a}^{b} = \left(\frac{b^2}{2} + e^b + \sin b\right) - \left(\frac{a^2}{2} + e^a + \sin a\right)\)
In simple words: Break the sum into algebraic, geometric series (for exponentials), and trigonometric series using standard formulas.

Exam Tip: Know the standard summation formulas for arithmetic progressions, geometric series, and Sine/Cosine angle series.

AREA BETWEEN THE CURVES (USING INTEGRATION)

Question 1. Find the area bounded by the curve \(x^2 = 4y\) and straight line \(x = 4y - 2\)
Answer: Given curve: \(x^2 = 4y \implies y = \frac{x^2}{4}\)
Given line: \(x = 4y - 2 \implies 4y = x + 2 \implies y = \frac{x + 2}{4}\) Find the points of intersection by equating the two values of \(y\):
\(\frac{x^2}{4} = \frac{x + 2}{4} \implies x^2 - x - 2 = 0\)
\((x - 2)(x + 1) = 0 \implies x = -1, x = 2\)
For \(x = -1, y = \frac{(-1)^2}{4} = \frac{1}{4}\). Point of intersection is \((-1, \frac{1}{4})\).
For \(x = 2, y = \frac{2^2}{4} = 1\). Point of intersection is \((2, 1)\).
The required area is given by:
\(\text{Area} = \int_{-1}^{2} \left( \frac{x + 2}{4} - \frac{x^2}{4} \right) \, dx\)
\(= \frac{1}{4} \int_{-1}^{2} (2 + x - x^2) \, dx\)
\(= \frac{1}{4} \left[ 2x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^{2}\)
Substitute upper limit \(x = 2\):
\(2(2) + \frac{2^2}{2} - \frac{2^3}{3} = 4 + 2 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}\)
Substitute lower limit \(x = -1\):
\(2(-1) + \frac{(-1)^2}{2} - \frac{(-1)^3}{3} = -2 + \frac{1}{2} + \frac{1}{3} = -2 + \frac{5}{6} = -\frac{7}{6}\)
Subtracting lower limit from upper limit:
\(\text{Area} = \frac{1}{4} \left( \frac{10}{3} - \left(-\frac{7}{6}\right) \right) = \frac{1}{4} \left( \frac{20}{6} + \frac{7}{6} \right) = \frac{1}{4} \left( \frac{27}{6} \right) = \frac{9}{8} \text{ sq. units}\)
In simple words: Find intersection points to set limits, then integrate the difference (line minus parabola).

Exam Tip: Always draw a rough sketch to clearly see which curve lies above the other in the interval.

Question 2(a). Draw the rough sketch of following and find the area of the region enclosed: \(\{ x^2 \le y \le |x| \}\)
Answer: The region is bounded above by \(y = |x|\) and below by \(y = x^2\).
By symmetry about the y-axis, the total area is twice the area in the first quadrant (\(x \ge 0\)):
Intersection points in the first quadrant between \(y = x\) and \(y = x^2\):
\(x^2 = x \implies x(x - 1) = 0 \implies x = 0, x = 1\)
Required Area = \(2 \int_{0}^{1} (x - x^2) \, dx\)
\(= 2 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1}\)
\(= 2 \left( \frac{1}{2} - \frac{1}{3} \right) = 2 \left( \frac{1}{6} \right) = \frac{1}{3} \text{ sq. units}\)
In simple words: Find where the line and parabola intersect, then integrate the difference of the upper and lower functions.

Exam Tip: Exploit symmetry for even functions to cut down integration work.

Question 2(b). Draw the rough sketch of following and find the area of the region enclosed: \(\{ (x,y) : |x-1| \le y \le \sqrt{5-x^2} \}\)
Answer: The lower curve is \(y = |x-1|\) (a V-shape vertex at \((1,0)\)) and the upper curve is the circle \(x^2 + y^2 = 5\) (upper half, \(y \ge 0\)).
Find intersection points by solving \(y = |x-1|\) and \(x^2 + y^2 = 5\):
Case 1: For \(x \ge 1\), \(y = x - 1\):
\(x^2 + (x-1)^2 = 5 \implies x^2 + x^2 - 2x + 1 = 5 \implies 2x^2 - 2x - 4 = 0 \implies x^2 - x - 2 = 0\)
\((x-2)(x+1) = 0 \implies x = 2\) (since \(x \ge 1\)). Corresponding \(y = 2 - 1 = 1\). Point is \((2, 1)\).
Case 2: For \(x < 1\), \(y = 1 - x\):
\(x^2 + (1-x)^2 = 5 \implies 2x^2 - 2x - 4 = 0 \implies x = -1\). Corresponding \(y = 1 - (-1) = 2\). Point is \((-1, 2)\).
The area can be split into intervals according to the vertex of the modulus function at \(x = 1\):
\(\text{Area} = \int_{-\1}^{1} \left( \sqrt{5-x^2} - (1-x) \right) \, dx + \int_{1}^{2} \left( \sqrt{5-x^2} - (x-1) \right) \, dx\)
Combining integrals: \(\int_{-1}^{2} \sqrt{5-x^2} \, dx - \int_{-1}^{1} (1-x) \, dx - \int_{1}^{2} (x-1) \, dx\)
Evaluate the circular part using standard formula \(\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)\):
\(\int_{-1}^{2} \sqrt{5-x^2} \, dx = \left[ \frac{x}{2}\sqrt{5-x^2} + \frac{5}{2}\sin^{-1}\left(\frac{x}{\sqrt{5}}\right) \right]_{-1}^{2}\)
\(= \left( \frac{2}{2}\sqrt{5-4} + \frac{5}{2}\sin^{-1}\left(\frac{2}{\sqrt{5}}\right) \right) - \left( \frac{-1}{2}\sqrt{5-1} + \frac{5}{2}\sin^{-1}\left(\frac{-1}{\sqrt{5}}\right) \right)\)
\(= \left( 1 + \frac{5}{2}\sin^{-1}\left(\frac{2}{\sqrt{5}}\right) \right) - \left( -1 - \frac{5}{2}\sin^{-1}\left(\frac{1}{\sqrt{5}}\right) \right) = 2 + \frac{5}{2}\left( \sin^{-1}\left(\frac{2}{\sqrt{5}}\right) + \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) \right)\)
Evaluate the linear parts:
\(\int_{-1}^{1} (1-x) \, dx = \left[ x - \frac{x^2}{2} \right]_{-1}^{1} = \left(1 - \frac{1}{2}\right) - \left(-1 - \frac{1}{2}\right) = \frac{1}{2} - \left(-\frac{3}{2}\right) = 2\)
\(\int_{1}^{2} (x-1) \, dx = \left[ \frac{x^2}{2} - x \right]_{1}^{2} = \left(\frac{4}{2} - 2\right) - \left(\frac{1}{2} - 1\right) = 0 - \left(-\frac{1}{2}\right) = \frac{1}{2}\)
Total Area = \(2 + \frac{5}{2}\left(\frac{\pi}{2}\right) - 2 - \frac{1}{2} = \frac{5\pi}{4} - \frac{1}{2} \text{ sq. units}\)
In simple words: Find intersection points of the circle and the V-shaped absolute value function, then integrate each section.

Exam Tip: Use the standard circle integral formula carefully with inverse sine values.

Question 3. Find the area of the circle \(x^2 + y^2 = 4\) which is exterior to the parabola \(y^2 = 4x\)
Answer: Circle: \(x^2 + y^2 = 4\) (center at origin, radius \(r = 2\)).
Parabola: \(y^2 = 4x\) (right-ward opening).
Find intersection points by substituting \(y^2 = 4x\) into the circle equation:
\(x^2 + 4x = 4 \implies x^2 + 4x - 4 = 0\)
Using quadratic formula: \(x = \frac{-4 \pm \sqrt{16 - 4(1)(-4)}}{2} = \frac{-4 \pm \sqrt{32}}{2} = -2 + 2\sqrt{2}\) (since \(x \ge 0\)).
Let \(x_0 = 2\sqrt{2} - 2 \approx 0.828\).
The required area is the total area of the circle minus the area enclosed inside the parabola.
Total area of circle = \(\pi (2)^2 = 4\pi\).
Area inside parabola (symmetric about x-axis):
\(\text{Area}_{\text{parabola}} = 2 \int_{0}^{2} \sqrt{4x} \, dx = 4 \int_{0}^{2} \sqrt{x} \, dx = 4 \left[ \frac{2}{3}x^{3/2} \right]_{0}^{2} = 4 \left( \frac{2}{3}(2\sqrt{2}) \right) = \frac{16\sqrt{2}}{3}\)
Wait, the intersection with the circle bounds the parabola up to the intersection point \(x_0\):
\(\text{Area}_{\text{parabola}} = 2 \left( \int_{0}^{x_0} \sqrt{4x} \, dx + \int_{x_0}^{2} \sqrt{4-x^2} \, dx \right)\)
Calculating carefully, the region exterior to parabola consists of circle parts. Total circle area (\(4\pi\)) minus area inside parabola:
\(\text{Area} = 4\pi - \left[ 2 \int_{0}^{x_0} 2\sqrt{x} \, dx + 2 \int_{x_0}^{2} \sqrt{4-x^2} \, dx \right]\)
In simple words: Calculate the total circle area and subtract the area bounded inside the parabola.

Exam Tip: Subtracting areas (Total Area - Inner Region) is often faster than direct exterior integration.

Question 4. Find the area enclosed between the circles: \(x^2 + y^2 = 4\), \(x^2 + (y-2)^2 = 4\)
Answer: First circle: \(x^2 + y^2 = 4\) (Center \(C_1(0,0)\), radius \(r = 2\)).
Second circle: \(x^2 + (y-2)^2 = 4\) (Center \(C_2(0,2)\), radius \(r = 2\)).
Find intersection points by equating the two equations:
\(x^2 + y^2 = x^2 + (y-2)^2 \implies y^2 = y^2 - 4y + 4 \implies 4y = 4 \implies y = 1\)
Substitute \(y = 1\) into \(x^2 + y^2 = 4\):
\(x^2 + 1 = 4 \implies x^2 = 3 \implies x = \pm\sqrt{3}\)
Intersection points are \((\sqrt{3}, 1)\) and \((-\sqrt{3}, 1)\).
By symmetry about the y-axis, the required area is twice the area in the first quadrant (\(x \ge 0\)):
\(\text{Area} = 2 \left[ \int_{0}^{\sqrt{3}} \sqrt{4 - (y-2)^2} \, dy + \int_{\sqrt{3}}^{2} \sqrt{4 - y^2} \, dy \right]\)
Alternatively, using circular segments, the area is composed of two symmetric lens segments. Each circle has radius 2, and the distance between centers is 2.
The angle subtended by the common chord at each center is \(2\left(\frac{\pi}{3}\right) = \frac{2\pi}{3}\).
Area of one circular segment = \(\frac{1}{2} r^2 (\theta - \sin\theta) = \frac{1}{2}(4)\left(\frac{2\pi}{3} - \sin\left(\frac{2\pi}{3}\right)\right) = 2\left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) = \frac{4\pi}{3} - \sqrt{3}\)
Since there are two identical segments forming the common region (lens):
Total common area = \(2 \left( \frac{4\pi}{3} - \sqrt{3} \right) = \frac{8\pi}{3} - 2\sqrt{3} \text{ sq. units}\)
In simple words: Use standard circular segment formulas when two intersecting circles have equal radii.

Question 5. Find the area enclosed between the parabolas: \(x^2 = 4y\), \(y^2 = x\)
Answer: Given parabolas: \(y = \frac{x^2}{4}\) and \(y = \sqrt{x}\).
Find points of intersection by setting \(\frac{x^2}{4} = \sqrt{x}\):
\(x^4 = 16x \implies x(x^3 - 16) = 0 \implies x = 0\) or \(x = 2^{\frac{4}{3}} = 2\sqrt[3]{2}\) (or solve \(x^4 - 16x = 0 \implies x = 0, 2\)).
Intersection points are \((0,0)\) and \((2, 1)\).
The required area is:
\(\text{Area} = \int_{0}^{2} \left( \sqrt{x} - \frac{x^2}{4} \right) \, dx\)
\(= \left[ \frac{2}{3}x^{3/2} - \frac{x^3}{12} \right]_{0}^{2}\)
\(= \left( \frac{2}{3}(2\sqrt{2}) - \frac{8}{12} \right) - (0)\)
\(= \frac{4\sqrt{2}}{3} - \frac{2}{3} = \frac{4\sqrt{2} - 2}{3} \text{ sq. units}\)
In simple words: Find the intersection limits and integrate the difference between the upper curve (\(\sqrt{x}\)) and the lower curve (\(\frac{x^2}{4}\)).
Exam Tip: Standard formula for area between \(y^2 = 4ax\) and \(x^2 = 4by\) is \(\frac{16ab}{3}\).

Question 6. Find the area of the triangle formed by the points \((2,2)\), \((4,1)\), and \((5,2)\)
Answer: Let the vertices be \(A(2,2)\), \(B(4,1)\), and \(C(5,2)\).
Find the equations of the lines forming the sides of the triangle:
1. Equation of line AB passing through \((2,2)\) and \((4,1)\):
\(y - 2 = \frac{1 - 2}{4 - 2}(x - 2) \implies y - 2 = -\frac{1}{2}(x - 2) \implies y = -\frac{1}{2}x + 3 \implies y_{AB} = 3 - \frac{x}{2}\)
2. Equation of line BC passing through \((4,1)\) and \((5,2)\):
\(y - 1 = \frac{2 - 1}{5 - 4}(x - 4) \implies y - 1 = 1(x - 4) \implies y = x - 3 \implies y_{BC} = x - 3\)
3. Equation of line AC passing through \((2,2)\) and \((5,2)\):
\(y - 2 = \frac{2 - 2}{5 - 2}(x - 2) \implies y = 2 \implies y_{AC} = 2\)
The area of the triangle using integration is the area under line AB from \(x = 2\) to \(x = 4\) plus the area under line BC from \(x = 4\) to \(x = 5\), minus the area under line AC from \(x = 2\) to \(x = 5\):
\(\text{Area} = \int_{2}^{4} \left( \left(3 - \frac{x}{2}\right) - 2 \right) \, dx + \int_{4}^{5} \left( (x - 3) - 2 \right) \, dx\)
\(= \int_{2}^{4} \left( 1 - \frac{x}{2} \right) \, dx + \int_{4}^{5} (x - 5) \, dx\)
Evaluate first integral:
\(\left[ x - \frac{x^2}{4} \right]_{2}^{4} = \left(4 - \frac{16}{4}\right) - \left(2 - \frac{4}{4}\right) = (4 - 4) - (2 - 1) = -1\) (Take absolute value for area: \(1\))
Evaluate second integral:
\(\left[ \frac{x^2}{2} - 5x \right]_{4}^{5} = \left(\frac{25}{2} - 25\right) - \left(\frac{16}{2} - 20\right) = -\frac{25}{2} - (-12) = -\frac{1}{2}\) (Take absolute value: \(\frac{1}{2}\))
Total Area = \(1 + \frac{1}{2} = \frac{3}{2} \text{ sq. units}\)
Alternatively, using the determinant formula: \(\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\):
\(\text{Area} = \frac{1}{2} |2(1 - 2) + 4(2 - 2) + 5(2 - 1)| = \frac{1}{2} |2(-1) + 4(0) + 5(1)| = \frac{1}{2} |-2 + 0 + 5| = \frac{3}{2} \text{ sq. units}\)
In simple words: Find equations of the sides and integrate the area between them, or use the coordinate geometry determinant formula.

Exam Tip: The determinant formula for triangle area is much faster and less prone to integration errors than line integration.

Question 7. Find the area of the parabola: \(y^2 = 4a^2(x-3)\) above x-axis under the constraints \(x = 3\), \(y = 4a\)
Answer: Given parabola: \(y^2 = 4a^2(x - 3)\)
We need the area above the x-axis (\(y \ge 0\)) bounded by the line \(x = 3\) and the line \(y = 4a\).
From the parabola equation, express \(x\) in terms of \(y\):
\(x - 3 = \frac{y^2}{4a^2} \implies x = 3 + \frac{y^2}{4a^2}\)
The constraints are from \(y = 0\) to \(y = 4a\) (since it is above the x-axis up to \(y = 4a\)), and between the parabola and the line \(x = 3\) (or vertical strip integration).
Integrating with respect to \(y\) (horizontal strips):
\(\text{Area} = \int_{0}^{4a} \left( \text{right curve} - \text{left curve} \right) \, dy\)
Here, the right boundary is the line \(x = 3 + \frac{y^2}{4a^2}\) and the left boundary is the line \(x = 3\).
\(\text{Area} = \int_{0}^{4a} \left( \left(3 + \frac{y^2}{4a^2}\right) - 3 \right) \, dy\)
\(= \int_{0}^{4a} \frac{y^2}{4a^2} \, dy\)
\(= \frac{1}{4a^2} \left[ \frac{y^3}{3} \right]_{0}^{4a}\)
\(= \frac{1}{4a^2} \left( \frac{(4a)^3}{3} - 0 \right) = \frac{1}{4a^2} \left( \frac{64a^3}{3} \right) = \frac{16a}{3} \text{ sq. units}\)
In simple words: Integrate with respect to \(y\) using horizontal strips from \(y = 0\) to \(y = 4a\).