RD Sharma Class 12 Ex 20.3 Solutions Chapter 20 Definite Integrals

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter20
Exercise20.3
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 20.3 Solutions Chapter 20 Definite Integrals

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Evaluate the following integrals:

Question 1(i). \int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}4x + 3 & , & \text{if }1 \leq x \leq 2 \\3x + 5 & , & \text{if }2 \leq x \leq 4\end{cases}

Solution:

We have,

I = \int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}4x + 3 & , & \text{if }1 \leq x \leq 2 \\3x + 5 & , & \text{if }2 \leq x \leq 4\end{cases}

I = \int_1^4 f\left( x \right) dx

Using additive property, we get

I = \int_1^2 f\left( x \right) d x + \int_2^4 f\left( x \right) d x

I = \int_1^2 \left( 4x + 3 \right) dx + \int_2^4 \left( 3x + 5 \right) dx

I = \left[ 2 x^2 + 3x \right]_1^2 + \left[ \frac{3 x^2}{2} + 5x \right]_2^4

I = 8 + 6 – 2 – 3 + 24 + 20 – 6 – 10

I = 37

Question 1(ii). \int\limits_0^9 f\left( x \right) dx, where\ f\left( x \right)= \begin{cases}\sin x & , & 0 \leq x \leq \frac{\pi}{2} \\ 1 & , & \frac{\pi}{2} \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}

Solution:

We have,

I = \int\limits_0^9 f\left( x \right) dx, where\ f\left( x \right)= \begin{cases}\sin x & , & 0 \leq x \leq \frac{\pi}{2} \\ 1 & , & \frac{\pi}{2} \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}

I = \int_0^9 f\left( x \right) d x

Using additive property, we get

I = \int_0^\frac{\pi}{2} f\left( x \right) d x + \int_\frac{\pi}{2}^3 f\left( x \right) d x + \int_3^9 f\left( x \right) d x

I = \int_0^\frac{\pi}{2} \sin x d x + \int_\frac{\pi}{2}^3 1 d x + \int_3^9 e^{x - 3} d x

I = \left[ - \cos x \right]_0^\frac{\pi}{2} + \left[ x \right]_\frac{\pi}{2}^3 + \left[ e^{x - 3} \right]_3^9

I = 0 + 1 + 3 – π/2 + e6 – e0 

I = 3 – π/2 + e6

Question 1(iii). \int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}

Solution:

We have,

I = \int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}

I = \int_1^4 f\left( x \right) d x

Using additive property, we get

I = \int_1^3 f\left( x \right) d x + \int_3^4 f\left( x \right) d x

I = \int_1^3 \left( 7x + 3 \right) d x + \int_3^4 8x d x

I = \left[ \frac{7 x^2}{2} + 3x \right]_1^3 + \left[ 4 x^2 \right]_3^4

I = 63/2 + 9 – 7/2 – 3 + 64 – 36

I = 56/2 + 34

I = 62

Question 2. \int\limits_{- 4}^4 \left| x + 2 \right| dx

Solution:

We have,

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

We know that, 

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

So, we get

I = \int_{- 4}^4 \left| x + 2 \right| d x

I = \int_{- 4}^{- 2} - \left( x + 2 \right) d x + \int_{- 2}^4 \left( x + 2 \right) d x

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

I = –2 + 4 – 8 – 8 + 8 + 8 – 2 + 4

I = 20

Question 3. \int\limits_{- 3}^3 \left| x + 1 \right| dx

Solution:

We have,

I = \int_{- 3}^3 \left| x + 1 \right| d x

We know that, 

\left| x + 1 \right| = \begin{cases} - \left( x + 1 \right) &, &- 3 \leq x \leq - 1 \\x + 1 &, &- 1 < x \leq 3\end{cases}

So, we get

I = \int_{- 3}^{- 1} - \left( x + 1 \right) d x + \int_{- 1}^3 \left[ x + 1 \right] d x

I = \left[ - \frac{\left( x + 1 \right)^2}{2} \right]_{- 3}^{- 1} + \left[ \frac{\left( x + 1 \right)^2}{2} \right]_{- 1}^3

I = 0 + 2 + 8 – 0

I = 10

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

Solution:

We have,

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

We know that,

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

So, we get

I = \int_{- 1}^\frac{- 1}{2} - \left( 2x + 1 \right) d x + \int_{- \frac{1}{2}}^1 \left( 2x + 1 \right) d x

I = - \left[ x^2 + x \right]_{- 1}^\frac{- 1}{2} + \left[ x^2 + x \right]_{- \frac{1}{2}}^1

I = –1/4 + 1/2 + 1 – 1 + 1 + 1 – 1/4 + 1/2

I = 5/2

Question 5. \int\limits_{- 2}^2 \left| 2x + 3 \right| dx

Solution:

We have,

I = \int\limits_{- 2}^2 \left| 2x + 3 \right| dx

We know that,

\left| 2x + 3 \right| = \begin{cases} - \left( 2x + 3 \right) &, &- 2 \leq x \leq - \frac{3}{2}\\\left( 2x + 3 \right)&, &- \frac{3}{2} < x \leq 2\end{cases}

So, we get

I = \int_{- 2}^\frac{- 3}{2} - \left( 2x + 3 \right) d x + \int_{- \frac{3}{2}}^2 \left( 2x + 3 \right) d x

I = - \left[ x^2 + 3x \right]_{- 2}^\frac{- 3}{2} + \left[ x^2 + 3x \right]_{- \frac{3}{2}}^2

I = –9/4 + 9/2 + 4 – 6 + 4 + 6 – 9/4 + 9/2

I = 25/2

Question 6. \int\limits_0^2 \left| x^2 - 3x + 2 \right| dx

Solution:

We have,

I = \int\limits_0^2 \left| x^2 - 3x + 2 \right| dx

We know that,

\left| x^2 - 3x + 2 \right| = \begin{cases} - \left( x^2 - 3x + 2 \right)&, &\left( x - 1 \right)\left( x - 2 \right) \leq 0 \text{ or}, 1 \leq x \leq 2\\\left( x^2 - 3x + 2 \right)&, &x^2 - 3x + 2 \geq 0 \text{ or}, x \in \left( - \infty , 1 \right) \cup \left( 2, \infty \right)\end{cases}

So we get,

I = \int_0^2 \left( x^2 - 3x + 2 \right) d x

I = \int_0^1 \left( x^2 - 3x + 2 \right) d x - \int_1^2 \left( x^2 - 3x + 2 \right) d x

I = \left[ \frac{x^3}{3} - \frac{3 x^2}{2} + 2x \right]_0^1 - \left[ \frac{x^3}{3} - \frac{3 x^2}{2} + 2x \right]_1^2

I = 1/3 – 3/2 + 2 – [8/3 – 6 + 4 – 1/3 + 3/2 – 2]

I = 1/3 – 3/2 + 2 – 8/3 + 6 – 2 + 1/3 – 3/2

I = 1

Question 7. \int\limits_0^3 \left| 3x - 1 \right| dx

Solution:

We have,

I = \int\limits_0^3 \left| 3x - 1 \right| dx

We know that,

\left| 3x - 1 \right| = \begin{cases} - \left( 3x - 1 \right)&,&0 \leq x \leq \frac{1}{3}\\\left( 3x - 1 \right)&,& \frac{1}{3} < x \leq 3\end{cases}

So we get,

I = \int_0^\frac{1}{3} - \left( 3x + 1 \right) dx + \int_\frac{1}{3}^0 \left( 3x + 1 \right) dx

I = \left[ \frac{- 3 x^2}{2} - x \right]_0^\frac{1}{3} + \left[ \frac{3 x^2}{2} + x \right]_\frac{1}{3}^3

I = –1/6 + 1/3 – 0 + 27/2 + 3 – 1/6 – 1/3

I = 65/6

Question 8. \int\limits_{- 6}^6 \left| x + 2 \right| dx

Solution:

We have,

I = \int\limits_{- 6}^6 \left| x + 2 \right| dx

We know that,

\left| x + 2 \right| = \begin{cases} - \left( x + 2 \right) &,& - 6 \leq x \leq - 2\\x + 2&,& - 2 < x \leq 6\end{cases}

So, we get 

I = \int_{- 6}^6 \left| x + 2 \right| d x

I = \int_{- 6}^{- 2} - \left( x + 2 \right) dx + \int_{- 2}^6 \left( x + 2 \right) dx

I = \left[ \frac{- x^2}{2} - 2x \right]_{- 6}^{- 2} + \left[ \frac{x^2}{2} + 2x \right]_{- 2}^6

I = –2 + 4 + 18 – 12 + 18 + 12 – 2 + 4

I = 40

Question 9. \int\limits_{- 2}^2 \left| x + 1 \right| dx

Solution:

We have,

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

We know that,

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

So we get,

I = \int_{- 2}^2 \left| x + 1 \right| d x

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

I = \left[ \frac{- x^2}{2} - x \right]_{- 2}^{- 1} + \left[ \frac{x^2}{2} + x \right]_{- 1}^2

I = -1/2 + 1 + 2 – 2 + 2 + 2 – 1/2 + 1

I = 5

Question 10. \int\limits_1^2 \left| x - 3 \right| dx

Solution:

We have,

I = \int\limits_1^2 \left| x - 3 \right| dx

We know that,

\left| x + 1 \right| = \begin{cases} - \left( x + 1 \right) &,& 1 \leq x \leq 3\\\left( x + 1 \right)&,& x > 3\end{cases}

So we get,

I = \int_1^2 \left| x - 3 \right| d x

I = \int_1^2 - \left( x - 3 \right) dx

I = \left[ \frac{- x^2}{2} - 3x \right]_1^2

I = – 2 – 6 + 1/2 + 3

I = – 5 + 1/2

I = (-10 + 1)/2

I = -9/2

Question 11. \int\limits_0^{\pi/2} \left| \cos 2x \right| dx

Solution:

We have,

I = \int\limits_0^{\pi/2} \left| \cos 2x \right| dx

We know that,

\left| \cos 2x \right| = \begin{cases} - \cos 2x &,& \frac{\pi}{4} \leq x \leq \frac{\pi}{2}\\\cos 2x&,& 0 < x \leq \frac{\pi}{4}\end{cases}

So we get,

I = \int_{- 2}^2 \left| \cos 2x \right| d x

I = \int_0^\frac{\pi}{4} \cos 2x dx - \int_\frac{\pi}{4}^\frac{\pi}{2} \cos 2x dx

I = \left[ \frac{\sin 2x}{2} \right]_0^\frac{\pi}{4} - \left[ \frac{\sin 2x}{2} \right]_\frac{\pi}{4}^\frac{\pi}{2}

I = 1/2 – 0 – 0 + 1/2

I = 1

Question 12. \int\limits_0^{2\pi} \left| \sin x \right| dx

Solution:

We have,

I = \int\limits_0^{2\pi} \left| \sin x \right| dx

We know that,

\left| \sin x \right| = \begin{cases} - \sin x &,& \pi \leq x \leq 2\pi\\\sin x&,& 0 < x \leq \pi\end{cases}

So we get,

I = \int_0^{2\pi} \left| \sin x \right| dx

I = \int_0^\pi \sin x dx + \int_\pi^{2\pi} - \sin x dx

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

I = 1 + 1 + 1 – (–1)

I = 1 + 1 + 1 + 1

I = 4

Question 13. \int\limits_{- \pi/4}^{\pi/4} \left| \sin x \right| dx

Solution:

We have,

I = \int\limits_{- \pi/4}^{\pi/4} \left| \sin x \right| dx

We know that,

\left| \sin x \right| = \begin{cases} - \sin x &,& - \frac{\pi}{4} \leq x \leq 0\\\sin x&,& 0 < x \leq \frac{\pi}{4}\end{cases}

So we get,

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

I = \int_{- \frac{\pi}{4}}^0 - \sin x dx + \int_0^\frac{\pi}{4} \sin x dx

I = \left[ \cos x \right]_\frac{- \pi}{4}^0 - \left[ \cos x \right]_0^\frac{- \pi}{4}

I = 1 – 1/√2 – 1/√2 + 1

I = 2 – 2/√2

I = 2 – √2

Question 14. \int\limits_2^8 \left| x - 5 \right| dx

Solution:

We have,

I = \int\limits_2^8 \left| x - 5 \right| dx

We know that,

\left| x - 5 \right| = \begin{cases} - \left( x - 5 \right) &,& 2 \leq x \leq 5\\x - 5&,& 5 < x \leq 8\end{cases}

So we get,

I = \int_2^8 \left| x - 5 \right| d x

I = \int_2^5 - \left( x - 5 \right) dx + \int_5^8 \left( x - 5 \right) dx

I = - \left[ \frac{x^2}{2} - 5x \right]_2^5 + \left[ \frac{x^2}{2} - 5x \right]_5^8

I = – 25/2 + 25 + 2 – 10 + 32 – 40 – 25/2 + 25

I = 9

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

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

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