Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.3 | Set 2

Read

Discuss

Evaluate the following integrals:

Question 15. $\int\limits_{- \pi/2}^{\pi/2} \left\{ \sin \left| x \right| + \cos \left| x \right| \right\} dx$

Solution:

We have,

I = $\int\limits_{- \pi/2}^{\pi/2} \left\{ \sin \left| x \right| + \cos \left| x \right| \right\} dx$

Since f(- x) = sin|- x| + cos|- x|

= sin |x| + cos |x|

= f(x)

So, f(x) is an even function.

Therefore, we get

I = $\int\limits_{- \pi/2}^{\pi/2} \left\{ \sin \left| x \right| + \cos \left| x \right| \right\} dx$

I = $2 \int_0^\frac{\pi}{2} \left( \sin x + \cos x \right) dx$

I = $2 \left[ - \cos x + \sin x \right]_0^\frac{\pi}{2}$

I = 2 (0 + 1 + 1 – 0)

I = 4

Question 16. $\int\limits_0^4 \left| x - 1 \right| dx$

Solution:

We have,

I = $\int\limits_0^4 \left| x - 1 \right| dx$

We know,

$\left| x - 1 \right| = \begin{cases} - \left( x - 1 \right) &,& 0 \leq x \leq 1\\x - 1&,& 1 < x \leq 4\end{cases}$

So we get,

I = $\int_0^4 \left| x - 1 \right| d x$

I = $\int_0^1 - \left( x - 1 \right) dx + \int_1^4 \left( x - 1 \right) dx$

I = $\left[ - \frac{x^2}{2} + x \right]_0^1 + \left[ \frac{x^2}{2} - x \right]_1^4$

I = -1/2 + 1 – 0 + 8 – 4 – 1/2 + 1

I = 5

Question 17. $\int\limits_1^4 \left\{ \left| x - 1 \right| + \left| x - 2 \right| + \left| x - 4 \right| \right\} dx$

Solution:

We have,

I = $\int\limits_1^4 \left\{ \left| x - 1 \right| + \left| x - 2 \right| + \left| x - 4 \right| \right\} dx$

We know,

$\left| x - 1 \right| = \begin{cases} - \left( x - 1 \right) &,& x \leq 1\\x - 1&,& 1 < x \leq 4\end{cases}$

$\left| x - 2 \right| = \begin{cases} - \left( x - 2 \right) &,& 1 \leq x \leq 2\\x - 2&,& 2 < x \leq 4\end{cases}$

$\left| x - 4 \right| = \begin{cases} - \left( x - 4 \right) &,& 1 \leq x \leq 4\\x - 4&,& x > 4\end{cases}$

So we get,

I = $\int_1^4 \left( x - 1 \right) d x - \int_1^2 \left( x - 2 \right) d x + \int_2^4 \left( x - 2 \right) d x - \int_1^4 \left( x - 4 \right) d x$

I = $\left[ \frac{x^2}{2} - x \right]_1^4 - \left[ \frac{x^2}{2} - 2x \right]_1^2 + \left[ \frac{x^2}{2} - 2x \right]_2^4 - \left[ \frac{x^2}{2} - 4x \right]_1^4$

I = 8 – 4 – 1/2 + 1 – (2 – 4 – 1/2 + 2) + 8 – 8 – 2 + 4 – (8 – 16 – 1/2 + 4)

I = 23/2

Question 18. $\int_{- 5}^0 \left\{ \left| x \right| + \left| x + 2 \right| + \left| x + 5 \right| \right\} dx$

Solution:

We have,

I = $\int_{- 5}^0 \left\{ \left| x \right| + \left| x + 2 \right| + \left| x + 5 \right| \right\} dx$

We know,

$\left| x \right| = \begin{cases} - x &,& - 5 \leq x \leq 0\\x&,& x > 0\end{cases}$

$\left| x + 2 \right| = \begin{cases} - \left( x + 2 \right) &,& - 5 \leq x \leq - 2\\x + 2&,& - 2 < x \leq 0\end{cases}$

$\left| x + 5 \right| = \begin{cases} - \left( x + 5 \right) &,& - 5 \leq x \leq 0\\x + 5&,& x > - 5\end{cases}$

So we get,

I = $- \int_{- 5}^0 x d x - \int_{- 5}^{- 2} \left( x + 2 \right) d x + \int_{- 2}^0 \left( x + 2 \right) d x + \int_{- 5}^0 \left( x + 5 \right) d x$

I = $- \left[ \frac{x^2}{2} \right]_{- 5}^0 - \left[ \frac{x^2}{2} + 2x \right]_{- 5}^{- 2} + \left[ \frac{x^2}{2} + 2x \right]_{- 2}^0 + \left[ \frac{x^2}{2} + 5x \right]_{- 5}^0$

I = 25/2 – (2 – 4 – 25/2 + 10) – 2 + 4 + (-25/2 + 25)

I = 63/2

Question 19. $\int\limits_0^4 \left( \left| x \right| + \left| x - 2 \right| + \left| x - 4 \right| \right) dx$

Solution:

We have,

I = $\int\limits_0^4 \left( \left| x \right| + \left| x - 2 \right| + \left| x - 4 \right| \right) dx$

We know,

$\left| x \right| = \begin{cases} - x &,& - 5 \leq x \leq 0\\x&,& x > 0\end{cases}$

$\left| x - 2 \right| = \begin{cases} - \left( x - 2 \right) &,& 0 \leq x \leq 2\\x - 2&,& 2 < x \leq 4\end{cases}$

$\left| x - 4 \right| = \begin{cases} - \left( x - 4 \right) &,& 0 \leq x \leq 4\\x - 4&,& x > 4\end{cases}$

So we get,

I = $\int_0^4 x d x - \int_0^2 \left( x - 2 \right) d x + \int_2^4 \left( x - 2 \right) d x - \int_0^4 \left( x - 4 \right) d x$

I = $\left[ \frac{x^2}{2} \right]_0^4 - \left[ \frac{x^2}{2} - 2x \right]_0^2 + \left[ \frac{x^2}{2} - 2x \right]_2^4 - \left[ \frac{x^2}{2} - 4x \right]_0^4$

I = 8 – (2 – 4) + 8 – 8 – 2 + 4 – (8 – 16)

I = 20

Question 20. $\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx$

Solution:

We have,

I = $\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx$

We know,

$\left| x + 1 \right| = \begin{cases}x + 1, & \text{if }x + 1 \geq 0 \\ - \left( x + 1 \right), & \text{if }x + 1 < 0\end{cases}$

When –1 < x < 0,

|x + 1| + |x| + |x – 1| = x + 1 + (- x) + [-(x – 1)]

= 2 – x

And when 0 < x < 1,

|x + 1| + |x| + |x – 1| = x + 1 + x + [-(x – 1)]

= x + 2

And when 1 ≤ x ≤ 2,

|x + 1| + |x| + |x – 1| = x + 1 + x + x – 1

= 3x

So we get,

I = $\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx$

I = $\int_{- 1}^0 \left( 2 - x \right)dx + \int_0^1 \left( x + 2 \right)dx + \int_1^2 3xdx$

I = $\left.\frac{\left( 2 - x \right)^2}{2 \times \left( - 1 \right)}\right|_{- 1}^0 + \left.\frac{\left( x + 2 \right)^2}{2}\right|_0^1 + \left.3 \times \frac{x^2}{2}\right|_1^2$

I = – 1/2(4 – 9) + 1/2( 9 – 4) + 3/2(4 – 1)

I = 5/2 + 5/2 + 9/2

I = 19/2

Question 21. $\int_{- 2}^2 x e^{\left| x \right|} dx$

Solution:

We have,

I = $\int_{- 2}^2 x e^{\left| x \right|} dx$

Now here,

f(- x) = (- x)e^{|- x|}

= – x e^|x|

= – f(x)

So, f(x) is an odd function.

Therefore we get,

I = $\int_{- 2}^2 x e^{\left| x \right|} dx$

I = 0

Question 22. $\int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx$

Solution:

We have,

I = $\int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx$

I = $\int_{- \frac{\pi}{4}}^0 \sin x\left| \sin x \right|dx + \int_0^\frac{\pi}{2} \sin x\left| \sin x \right|dx$

As we know, $\left| \sin x \right| = \begin{cases}\sin x, & 0 \leq x \leq \frac{\pi}{2} \\ - \sin x, & - \frac{\pi}{4} \leq x \leq 0\end{cases}$

I = $\int_{- \frac{\pi}{4}}^0 \sin x\left( - \sin x \right)dx + \int_0^\frac{\pi}{2} \sin x\sin xdx$

I = $- \int_{- \frac{\pi}{4}}^0 \sin^2 xdx + \int_0^\frac{\pi}{2} \sin^2 xdx$

I = $- \int_{- \frac{\pi}{4}}^0 \frac{1 - \cos2x}{2}dx + \int_0^\frac{\pi}{2} \frac{1 - \cos2x}{2}dx$

I = $- \frac{1}{2} \int_{- \frac{\pi}{4}}^0 dx + \frac{1}{2} \int_{- \frac{\pi}{4}}^0 \cos2xdx + \frac{1}{2} \int_0^\frac{\pi}{2} dx - \frac{1}{2} \int_0^\frac{\pi}{2} \cos2xdx$

I = $\left.- \frac{1}{2} \times x\right|_{- \frac{\pi}{4}}^0 +\left. \frac{1}{2} \times \frac{\sin2x}{2}\right|_{- \frac{\pi}{4}}^0 + \left.\frac{1}{2} \times x\right|_0^\frac{\pi}{2} - \left.\frac{1}{2} \times \frac{\sin2x}{2}\right|_0^\frac{\pi}{2}$

I = -1/2(0 + π/4) + 1/4(0 + sin π/2) + 1/2 ( π/2 – 0) – 1/4(sin π – 0)

I = – π/8 + 1/4 (0 + 1) + π/8 – 1/4 (0 – 0)

I = π/8 + 1/4

Question 23. $\int_0^\pi \cos x\left| \cos x \right|dx$

Solution:

We have,

I = $\int_0^\pi \cos x\left| \cos x \right|dx$

Now here,

f(π – x) = cos(π – x)|cos(π – x)|

= -cos x|-cos x|

= – cos x|cos x|

= – f(x)

So, f(x) is an odd function.

Therefore we get,

I = $\int_0^\pi \cos x\left| \cos x \right|dx$

I = 0

Question 24. $\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx$

Solution:

We have,

I = $\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx$

Now here,

f(- x) = 2sin|- x| + cos|- x|

= 2sin|x| + cos|x|

= f(x)

So, f(x) is an odd function.

Therefore we get,

I = $\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx$

I = $2 \int_0^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx$

As we know, $\left| x \right| = \begin{cases}x, & \text{if }x \geq 0 \\ - x, & \text{if }x < 0\end{cases}$

I = $2 \int_0^\frac{\pi}{2} \left( 2\sin x + \cos x \right)dx$

I = $4 \int_0^\frac{\pi}{2} \sin x\ dx + 2 \int_0^\frac{\pi}{2} \cos x\ dx$

I = $\left.4 \times \left( - \cos x \right)\right|_0^\frac{\pi}{2} + \left.2 \times \sin x\right|_0^\frac{\pi}{2}$

I = – 4(cos π/2 – cos 0) + 2(sin π/2 – sin 0)

I = –4 ( 0 – 1) + 2 (1 – 0)

I = 4 + 2

I = 6

Question 25. $\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx$

Solution:

We have,

I = $\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx$

I = $\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin^{- 1} \left( \sin x \right)dx + \int_\frac{\pi}{2}^\pi \sin^{- 1} \left( \sin x \right)dx$

I = $\int_{- \frac{\pi}{2}}^\frac{\pi}{2} xdx + \int_\frac{\pi}{2}^\pi \left( \pi - x \right)dx$

As π/2 ≤ x ≤ π, we get

=> –π ≤ –x ≤ –π/2

=> 0 ≤ π – x ≤ π/2

So, we get

I = $\left.\frac{x^2}{2}\right|_{- \frac{\pi}{2}}^\frac{\pi}{2} + \left.\frac{\left( \pi - x \right)^2}{2 \times \left( - 1 \right)}\right|_\frac{\pi}{2}^\pi$

I = 1/2 (π²/4 – π²/4) – 1/2( 0 – π²/4)

I = 0 + π²/8

I = π²/8

Question 26. $\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx$

Solution:

We have,

I = $\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx$

I = $- \frac{\pi}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{1}{\sqrt{\cos x \sin^2 x}}dx$

I = $- \frac{\pi}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{1}{\sqrt{\cos x}\left| \sin x \right|}dx$

As we know, $f\left( - x \right) = \sqrt{\cos\left( - x \right)}\left| \sin\left( - x \right) \right|$

= √cos x|-sin x|

= √cos x|sin x|

= f(x)

So, f(x) is an odd function.

Therefore we get,

I = $- \frac{\pi}{2} \times 2 \int_0^\frac{\pi}{2} \frac{1}{\sqrt{\cos x}\left| \sin x \right|}dx$

I = $-\pi\int_0^\frac{\pi}{2} \frac{1}{\sqrt{\cos x}\left| \sin x \right|}dx$

As we know, $\left| \sin x \right| = \sin x, 0 \leq x \leq \frac{\pi}{2}$ ,

I = $- \pi \int_0^\frac{\pi}{2} \frac{1}{\sqrt{\cos x}\sin x}dx$

I = $- \pi \int_0^\frac{\pi}{2} \frac{\sin x}{\sqrt{\cos x}\left( 1 - \cos^2 x \right)}dx$

Let cos x = z². So, we have

=> – sin x dx = 2z dz

Now, the lower limit is, x = 0

=> z² = cos x

=> z² = cos 0

=> z² = 1

=> z = 1

Also, the upper limit is, x = π/2

=> z² = cos x

=> z² = cos π/2

=> z² = 0

=> z = 0

So, the equation becomes,

I = $2\pi \int_1^0 \frac{zdz}{z\left( 1 - z^4 \right)}$

I = $2\pi \int_1^0 \frac{dz}{1 - z^4}$

I = $2\pi \int_1^0 \frac{dz}{\left( 1 - z \right)\left( 1 + z \right)\left( 1 + z^2 \right)}$

I = $2\pi \int_1^0 \frac{\frac{1}{4}}{1 - z}dz + 2\pi \int_1^0 \frac{\frac{1}{4}}{1 + z}dz + 2\pi \int_1^0 \frac{\frac{1}{2}}{1 + z^2}dz$

I = $\left.\frac{2\pi}{4} \times \frac{\log\left( 1 - z \right)}{- 1}\right|_1^0 + \left.\frac{2\pi}{4} \times \log\left( 1 + z \right)\right|_1^0 + \left.\frac{2\pi}{2} \times \tan^{- 1} z\right|_1^0$

I = – π/2(log1 – log0) + π/2(log1 – log2) + π(tan^-1 0 – tan^-1 1)

I = – π/2[0 – ∞] + π/2(0 – log2) + π(0 – π/4)

I = -∞ – π/2 log2 – π²/4

I = –∞

Question 27. $\int_0^2 2x\left[ x \right]dx$

Solution:

We have,

I = $\int_0^2 2x\left[ x \right]dx$

I = $\int_0^1 2x\left[ x \right]dx + \int_1^2 2x\left[ x \right]dx$

As we know, $\left[ x \right] = \begin{cases}0, & 0 \leq x < 1 \\ 1, & 1 \leq x < 2\end{cases}$

I = $\int_0^1 2x \times 0dx + \int_1^2 2x \times 1dx$

I = $0 + 2 \int_1^2 xdx$

I = $\left.2 \times \frac{x^2}{2}\right|_1^2$

I = 4 – 1

I = 3

Question 28. $\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx$

Solution:

We have,

I = $\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx$

I = $\int_0^\pi \cos^{- 1} \left( \cos x \right)dx +\int_\pi^{2\pi} \cos^{- 1} \left( \cos x \right)dx$

As we know, π ≤ x ≤ 2π

=> –2π ≤ –x ≤ –π

=> 0 ≤ 2π – x ≤ π

Therefore, we get

I = $\int_0^\pi xdx + \int_\pi^{2\pi} \left( 2\pi - x \right)dx$

I = $\left.\frac{x^2}{2}\right|_0^\pi + \left.\frac{\left( 2\pi - x \right)^2}{2 \times \left( - 1 \right)}\right|_\pi^{2\pi}$

I = 1/2( π – 0) – 1/2(0 – π)

I = π²/2 + π²/2

I = π²

My Personal Notes

Last Updated : 14 Jul, 2021

Related Articles

Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.3 | Set 2

Evaluate the following integrals:

Question 15. $\int\limits_{- \pi/2}^{\pi/2} \left\{ \sin \left| x \right| + \cos \left| x \right| \right\} dx$

Question 16. $\int\limits_0^4 \left| x - 1 \right| dx$

Question 17. $\int\limits_1^4 \left\{ \left| x - 1 \right| + \left| x - 2 \right| + \left| x - 4 \right| \right\} dx$

Question 18. $\int_{- 5}^0 \left\{ \left| x \right| + \left| x + 2 \right| + \left| x + 5 \right| \right\} dx$

Question 19. $\int\limits_0^4 \left( \left| x \right| + \left| x - 2 \right| + \left| x - 4 \right| \right) dx$

Question 20. $\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx$

Question 21. $\int_{- 2}^2 x e^{\left| x \right|} dx$

Question 22. $\int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx$

Question 23. $\int_0^\pi \cos x\left| \cos x \right|dx$

Question 24. $\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx$

Question 25. $\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx$

Question 26. $\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx$

Question 27. $\int_0^2 2x\left[ x \right]dx$

Question 28. $\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx$

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Related Articles

Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.3 | Set 2

Evaluate the following integrals:

Question 15.

Question 16.

Question 17.

Question 18.

Question 19.

Question 20.

Question 21.

Question 22.

Question 23.

Question 24.

Question 25.

Question 26.

Question 27.

Question 28.

Please Login to comment...

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Question 15. $\int\limits_{- \pi/2}^{\pi/2} \left\{ \sin \left| x \right| + \cos \left| x \right| \right\} dx$

Question 16. $\int\limits_0^4 \left| x - 1 \right| dx$

Question 17. $\int\limits_1^4 \left\{ \left| x - 1 \right| + \left| x - 2 \right| + \left| x - 4 \right| \right\} dx$

Question 18. $\int_{- 5}^0 \left\{ \left| x \right| + \left| x + 2 \right| + \left| x + 5 \right| \right\} dx$

Question 19. $\int\limits_0^4 \left( \left| x \right| + \left| x - 2 \right| + \left| x - 4 \right| \right) dx$

Question 20. $\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx$

Question 21. $\int_{- 2}^2 x e^{\left| x \right|} dx$

Question 22. $\int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx$

Question 23. $\int_0^\pi \cos x\left| \cos x \right|dx$

Question 24. $\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx$

Question 25. $\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx$

Question 26. $\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx$

Question 27. $\int_0^2 2x\left[ x \right]dx$

Question 28. $\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx$