CBSE Class 12 Mathematics Indefinite And Definite Integrals Worksheet Set 04

CBSE Class 12 Mathematics Indefinite and Definite Integrals (4). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. Evaluate:
\[ I = \int_{-1}^{2} |x^3 - x| \, dx \]
Answer:
\[ I = \int_{-1}^{2} |x(x+1)(x-1)| \, dx \]
The critical points are at \( x = 0, -1, 1 \).
\[ \dots I = + \int_{-1}^{0} (x^3 - x) \, dx - \int_{0}^{1} (x^3 - x) \, dx + \int_{1}^{2} (x^3 - x) \, dx \]
\[ I = \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{-1}^{0} - \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{0}^{1} + \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{1}^{2} \]
\[ I = \left[ (0) - \left( \frac{1}{4} - \frac{1}{2} \right) \right] - \left[ \left( \frac{1}{4} - \frac{1}{2} \right) - (0) \right] + \left[ (4 - 2) - \left( \frac{1}{4} - \frac{1}{2} \right) \right] \]
\[ I = \frac{11}{4} \quad \text{ans.} \]

Question. Evaluate:
\[ I = \int_{1}^{4} (|x - 3| + |x - 1| + |x - 2|) \, dx \]
Answer:
We can rewrite the integral by ordering the terms:
\[ I = \int_{1}^{4} (|x - 1| + |x - 2| + |x - 3|) \, dx \]
The critical points are \( x = 1, 2, 3 \).
\[ I = \int_{1}^{2} [ (x - 1) - (x - 2) - (x - 3) ] \, dx + \int_{2}^{3} [ (x - 1) + (x - 2) - (x - 3) ] \, dx + \int_{3}^{4} [ (x - 1) + (x - 2) + (x - 3) ] \, dx \]
\[ I = \int_{1}^{2} (-x + 4) \, dx + \int_{2}^{3} x \, dx + \int_{3}^{4} (3x - 6) \, dx \]
\[ I = \left[ \frac{-x^2}{2} + 4x \right]_{1}^{2} + \left[ \frac{x^2}{2} \right]_{2}^{3} + \left[ \frac{3x^2}{2} - 6x \right]_{3}^{4} \]
\[ I = \left[ (-2 + 8) - \left( -\frac{1}{2} + 4 \right) \right] + \left[ \frac{9}{2} - 2 \right] + \left[ (24 - 24) - \left( \frac{27}{2} - 18 \right) \right] \]
\[ I = 6 - \frac{7}{2} + \frac{5}{2} + \frac{9}{2} \]
\[ I = \frac{19}{2} \quad \text{ans.} \]

Question. Evaluate:
\[ I = \int_{0}^{2\pi} |\sin x| \, dx \]
Answer:
\[ I = \int_{0}^{\pi} \sin x \, dx - \int_{\pi}^{2\pi} \sin x \, dx \]
\[ I = [-\cos x]_{0}^{\pi} - [-\cos x]_{\pi}^{2\pi} \]
\[ I = -[\cos x]_{0}^{\pi} + [\cos x]_{\pi}^{2\pi} \]
\[ I = -[\cos \pi - \cos 0] + [\cos 2\pi - \cos \pi] \]
\[ I = -[-1 - 1] + [1 - (-1)] \]
\[ I = 2 + 2 \]
\[ I = 4 \quad \text{ans.} \]

Question. Evaluate:
\[ I = \int_{0}^{\frac{\pi}{2}} |\cos(2x)| \, dx \]
Answer:
Critical point: \( 2x = \frac{\pi}{2} \implies x = \frac{\pi}{4} \)
\[ \therefore I = \int_{0}^{\frac{\pi}{4}} \cos(2x) \, dx - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos(2x) \, dx \]
Since:
For \( 0 < x < \frac{\pi}{4} \implies 0 < 2x < \frac{\pi}{2} \) (1st quad, positive)
For \( \frac{\pi}{4} < x < \frac{\pi}{2} \implies \frac{\pi}{2} < 2x < \pi \) (2nd quad, negative)
\[ I = \frac{1}{2} [\sin(2x)]_{0}^{\frac{\pi}{4}} - \frac{1}{2} [\sin(2x)]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \]
\[ I = \frac{1}{2} \left[ \sin\left(\frac{\pi}{2}\right) - \sin 0 \right] - \frac{1}{2} \left[ \sin \pi - \sin\left(\frac{\pi}{2}\right) \right] \]
\[ I = \frac{1}{2} [1 - 0] - \frac{1}{2} [0 - 1] \]
\[ I = \frac{1}{2} + \frac{1}{2} \]
\[ I = 1 \quad \text{ans.} \]

Question. Evaluate:
\[ I = \int_{0}^{3} [x] \, dx \quad \text{(greatest integer function)} \]
Answer:
\[ I = \int_{0}^{1} [x] \, dx + \int_{1}^{2} [x] \, dx + \int_{2}^{3} [x] \, dx \]
\[ I = \int_{0}^{1} (0) \, dx + \int_{1}^{2} (1) \, dx + \int_{2}^{3} (2) \, dx \]
\[ I = 0 + (x)_{1}^{2} + (2x)_{2}^{3} \]
\[ I = (2 - 1) + (6 - 4) \]
\[ I = 3 \quad \text{ans.} \]

Question. Evaluate:
\[ I = \int_{0}^{2} [x^2] \, dx \quad \text{(greatest integer function)} \]
Answer:
\[ I = \int_{0}^{2} [x^2] \, dx \]
The critical points of \( [x^2] \) inside \( [0, 2] \) are \( x = 1, \sqrt{2}, \sqrt{3} \).
\[ I = \int_{0}^{1} [x^2] \, dx + \int_{1}^{\sqrt{2}} [x^2] \, dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^2] \, dx + \int_{\sqrt{3}}^{2} [x^2] \, dx \]
\[ I = \int_{0}^{1} (0) \, dx + \int_{1}^{\sqrt{2}} (1) \, dx + \int_{\sqrt{2}}^{\sqrt{3}} (2) \, dx + \int_{\sqrt{3}}^{2} (3) \, dx \]
\[ I = 0 + (x)_{1}^{\sqrt{2}} + (2x)_{\sqrt{2}}^{\sqrt{3}} + (3x)_{\sqrt{3}}^{2} \]
\[ I = (\sqrt{2} - 1) + (2\sqrt{3} - 2\sqrt{2}) + (6 - 3\sqrt{3}) \]
\[ I = 5 - \sqrt{2} - \sqrt{3} \quad \text{Ans...} \]

Question. Evaluate:
\[ I = \int_{0}^{1.5} [x^2] \, dx \quad \text{(greatest integer function)} \]
Answer:
\[ I = \int_{0}^{1.5} [x^2] \, dx \]
The critical points are at \( x = 1 \) and \( x = \sqrt{2} \approx 1.414 \).
\[ I = \int_{0}^{1} [x^2] \, dx + \int_{1}^{\sqrt{2}} [x^2] \, dx + \int_{\sqrt{2}}^{1.5} [x^2] \, dx \]
\[ I = \int_{0}^{1} 0 \, dx + \int_{1}^{\sqrt{2}} (1) \, dx + \int_{\sqrt{2}}^{1.5} (2) \, dx \]
\[ I = 0 + (x)_{1}^{\sqrt{2}} + (2x)_{\sqrt{2}}^{1.5} \]
\[ I = (\sqrt{2} - 1) + (3 - 2\sqrt{2}) \]
\[ I = 2 - \sqrt{2} \quad \text{ans.} \]

Question. Evaluate:
\[ I = \int_{-1}^{\frac{3}{2}} |x \sin(\pi x)| \, dx \]
Answer:
Case - I:
\( -1 < x < 0 \implies -\pi < \pi x < 0 \)
\( |x \sin(\pi x)| = x \sin(\pi x) \)

Case - II:
\( 0 < x < 1 \implies 0 < \pi x < \pi \)
\( |x \sin(\pi x)| = x \sin(\pi x) \)

Case - III:
\( 1 < x < \frac{3}{2} \implies \pi < \pi x < \frac{3\pi}{2} \)
\( |x \sin(\pi x)| = -x \sin(\pi x) \)

\[ \therefore I = \int_{-1}^{1} x \sin(\pi x) \, dx - \int_{1}^{\frac{3}{2}} x \sin(\pi x) \, dx \]

Let \( I_1 = \int x \sin(\pi x) \, dx \)
Using integration by parts:
\[ I_1 = \left[ \frac{-x \cos(\pi x)}{\pi} \right] - \int (1) \frac{(-\cos(\pi x))}{\pi} \, dx \]
\[ I_1 = \frac{-x \cos(\pi x)}{\pi} + \frac{1}{\pi} \int \cos(\pi x) \, dx \]
\[ I_1 = \frac{-x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \]

Substituting the integrated function back:
\[ \therefore I = \left[ \frac{-x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{-1}^{1} - \left[ \frac{-x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2} \right]_{1}^{3/2} \]
\[ I = \left[ \left( \frac{-\cos(\pi)}{\pi} + \frac{\sin(\pi)}{\pi^2} \right) - \left( \frac{\cos(-\pi)}{\pi} + \frac{\sin(-\pi)}{\pi^2} \right) \right] - \left[ \left( \frac{-\frac{3}{2} \cos\left(\frac{3\pi}{2}\right)}{\pi} + \frac{\sin\left(\frac{3\pi}{2}\right)}{\pi^2} \right) - \left( \frac{-\cos(\pi)}{\pi} + \frac{\sin(\pi)}{\pi^2} \right) \right] \]
Evaluating using trig values:
\[ I = \left[ \left( \frac{1}{\pi} + 0 \right) - \left( \frac{-1}{\pi} + 0 \right) \right] - \left[ \left( 0 - \frac{1}{\pi^2} \right) - \left( \frac{1}{\pi} + 0 \right) \right] \]
\[ \dots \left\{ \begin{aligned} &\cos(\pi) = -1 \;; \quad \cos(-\pi) = -1 \\ &\sin(-\pi) = 0 \;; \quad \sin\left(\frac{3\pi}{2}\right) = -1 \\ &\cos(3\pi/2) = 0 \end{aligned} \right\} \]
\[ I = \frac{1}{\pi} + \frac{1}{\pi} + \frac{1}{\pi^2} + \frac{1}{\pi} \]
\[ \therefore I = \frac{3}{\pi} + \frac{1}{\pi^2} \quad \text{ans.} \]

Question. Evaluate:
\[ I = \int_{1/e}^{e} |\log x| \, dx \]
Answer:
\[ I = \int_{1/e}^{e} |\log x| \, dx \]
The critical point is \( x = 1 \) since \( \log 1 = 0 \).
\[ \therefore I = -\int_{1/e}^{1} \log x \, dx + \int_{1}^{e} \log x \, dx \]
Let \( I_1 = \int \log x \, dx \)
\[ I_1 = \int \log x \cdot 1 \, dx \]
\[ I_1 = \log x \cdot x - \int \frac{1}{x} \cdot x \, dx \]
\[ I_1 = x \log x - x \]

Evaluating the limits:
\[ \therefore I = -[x \log x - x]_{1/e}^{1} + [x \log x - x]_{1}^{e} \]
\[ I = - \left[ (\log 1 - 1) - \left( \frac{1}{e} \log\left(\frac{1}{e}\right) - \frac{1}{e} \right) \right] + [(e \log e - e) - (\log 1 - 1)] \]
\[ I = - \left[ (0 - 1) - \left( \frac{-1}{e} - \frac{1}{e} \right) \right] + [(e - e) - (0 - 1)] \quad \dots \left\{\log e = 1, \; \log\left(\frac{1}{e}\right) = -1 \right\} \]
\[ I = 1 - \frac{2}{e} + 1 \]
\[ I = 2 - \frac{2}{e} \quad \text{ans.} \]

Question. Evaluate:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^7 x \, dx \]
Answer:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^7 x \, dx \]
Here,
\[ f(x) = \sin^7 x \]
\[ f(-x) = \sin^7(-x) = (-\sin x)^7 = -f(x) \]
\[ \therefore f(-x) = -f(x) \]
So \( f(x) \) is an odd function.
\[ \therefore I = 0 \quad \text{Ans.} \quad \dots \left\{ \int_{-a}^{a} f(x) \, dx = 0 \text{ when } f(x) \text{ is an odd function} \right\} \]

 Please click the link below to download CBSE Class 12 Mathematics Indefinite and Definite Integrals (4).