NCERT Solutions of class 12 maths Miscellaneous exercise -Application of Integrals

The NCERT solutions of miscellaneous exercise of chapter 8-Application of Integrals are a summary of all the methods used in all the exercises of chapter 8-Application of Integrals, therefore in this respect, NCERT solutions of miscellaneous exercise are very important for students.All the questions of miscellaneous exercise are solved by the expert by a step-by-step method so each student will understand the solution clearly.

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NCERT Solutions of class 12 maths Miscellaneous exercise -Application of Integrals

Exercise 8.1-Application of Integration

Exercise 8.2-Application of Integration

Exercise 8.3-Application of Integration

NCERT Solutions of Maths for Class 12 Miscellaneous exercise -Application of Integrals

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Q1. Find the area under the given curve and given lines

(i) y = x² , x = 1, x =2 and x-axis

(ii) y = x⁴, x=1, x= 5 and x-axis

Ans.

(i) We are given the curve y = x² and the lines x = 1, x =2

Area covered by the curve,x-axis and the given lines = arABCD

$\boldsymbol{arABCD = \int_{1}^{2}y.dx}$

$\boldsymbol{ = \int_{1}^{2}x^{2}.dx}$

$\boldsymbol{=\left [ \frac{x^{3}}{3} \right ]_{1}^{2}}$

$\boldsymbol{=\frac{1}{3}\left ( 2^{3} -1^{3}\right )}$

$\boldsymbol{=\frac{1}{3}\left ( 8 -1\right )}$

$\boldsymbol{=\frac{7}{3}}$

(ii) We are given the curve y = x⁴, x=1, x= 5 and the lines x = 1, x =2

Area covered by the curve, x-axis and the given lines = arABCD

$\boldsymbol{arABCD= \int_{1}^{5}y.dx}$

$\boldsymbol{= \int_{1}^{5}x^{4}.dx}$

$\boldsymbol{=\left [ \frac{x^{5}}{5} \right ]_{1}^{5}}$

$\boldsymbol{=\frac{1}{5}\left ( 5^{5} -1\right )}$

$\boldsymbol{=\frac{3125-1}{5}=\frac{3124}{5}\: sq.unit}$

Q2. Find the area between the curves y = x and y = x²

Ans. We are given the equations of curve y = x² and y = x

Putting y = x, in y = x²

x = x² ⇒ x= 0 and x =1,again putting x = 0 and x= 1 in one of the equations, we get y=0 and y=1

Therefore both of curve and the line intersect each other at origin (0,0) and (1,1)

Area covered by the line and the curve , area OCB = ar ΔOAB – ar OABCO

$\boldsymbol{arOABCO=\int_{0}^{1}y.dx}$

$\boldsymbol{ =\int_{0}^{1}x^{2}.dx}$

$\boldsymbol{=\left [ \frac{x^{3}}{3} \right ]_{0}^{1}}$

$\boldsymbol{=\frac{1}{3}\left ( 1-0 \right )=\frac{1}{3}}$

ar OABCO = 1/3 sq.unit

OA = 1 unit and AB = 1 unit

$\boldsymbol{ar\Delta OAB = \frac{1}{2}\times 1\times 1=\frac{1}{2}\: sq.unit}$

$\boldsymbol{arOCB=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\: sq.unit}$

Q3. Find the area of the region lying in the first quadrant and bounded by y = 4x², x = 0, y =1 and y = 4.

Ans. The given curve is y = 4x² and the lines are x = 0, y =1 and y = 4.

The area covered by the curve y = 4x² and the lines are x = 0(y-axis), y =1, and y = 4 is ar ABCD

$\boldsymbol{arABCD =\int_{1}^{4}x.dy}$

From the equation of curve, y = 4x², we have $\boldsymbol{x =\frac{\sqrt{y}}{2}}$

$\boldsymbol{=\int_{1}^{4}\frac{\sqrt{y}}{2}.dy}$

$\boldsymbol{=\frac{1}{2}\left [\frac{y^{3/2}}{3/2} \right ]_{1}^{4}}$

$\boldsymbol{=\frac{1}{3}\left (4^{3/2} -1^{3/2}\right )}$

$\boldsymbol{=\frac{1}{3}\left ( 2^{3} -1\right )=\frac{1}{3}\left ( 8-1 \right )=\frac{7}{3}\: sq.unit}$

Q4. Sketch the graph of $\boldsymbol{y = \left | x+3 \right |}$ and evaluate $\boldsymbol{\int_{-6}^{1}\left | x+3 \right |.dx}$

rom the equation of curve, y = 4x², we have $\boldsymbol{x =\frac{\sqrt{y}}{2}}$

$\boldsymbol{=\int_{1}^{4}\frac{\sqrt{y}}{2}.dy}$

$\boldsymbol{=\frac{1}{2}\left [\frac{y^{3/2}}{3/2} \right ]_{1}^{4}}$

$\boldsymbol{=\frac{1}{3}\left (4^{3/2} -1^{3/2}\right )}$

$\boldsymbol{=\frac{1}{3}\left ( 2^{3} -1\right )=\frac{1}{3}\left ( 8-1 \right )=\frac{7}{3}\: sq.unit}$

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Q4. Sketch the graph of $\boldsymbol{y = \left | x+3 \right |}$ and evaluate $\boldsymbol{\int_{-6}^{1}\left | x+3 \right |.dx}$

Ans.Solutions of the given equation $\boldsymbol{y = \left | x+3 \right |}$

x	-6	-5	-4	-3	-2	-1	0
y	3	2	1	0	1	2	3

The graph of the given equation is drawn as follows

$\boldsymbol{\int_{-6}^{1}\left | x+3 \right |.dx}$

$\boldsymbol{=-\int_{-6}^{-3}\left ( x+3 \right ).dx+\int_{-3}^{0}\left ( x+3 \right ).dx}$

$\boldsymbol{=-\left [ \frac{x^{2}}{2}+3x \right ]_{-6}^{-3}+\left [ \frac{x^{2}}{2} +3x\right ]_{-3}^{0}}$

$\boldsymbol{=-\left [ \frac{\left ( -3 \right )^{2}}{2}+3\times-3-\frac{\left ( -6 \right )^{2}}{2}-3\times -6 \right ]+\left [ 0-\frac{\left ( -3 \right )^{2}}{2}- 3\times -3\right ]}$

$\boldsymbol{=-\frac{9}{2}+9+18-18-\frac{9}{2}+9}$

$\boldsymbol{=-\frac{18}{2}+18}$

= – 9 + 18 = 9

Q5. Find the area bounded by the curve y = sinx between x = 0 and x = 2π.

Ans.The graph of the curve y = sinx between x = 0 and x = 2π, is shown below

The area bounded by the curve y = sinx, x = 0 and x = 2π

=Area of OAB + area of BCD

$\boldsymbol{=\int_{0}^{\pi }y.dx+\int_{\pi }^{2\pi }y.dx}$

$\boldsymbol{=\int_{0}^{\pi }sinx.dx+\int_{\pi }^{2\pi }sinx.dx}$

$\boldsymbol{=\left [ -cosx \right ]_{0}^{\pi }+\left [ -cosx \right ]_{\pi }^{2\pi }}$

$\boldsymbol{=\left ( -cos\pi +cos0 \right )+\left | -cos2\pi +cos\pi \right |}$

$\boldsymbol{=1+1+\left | -1-1 \right |}$

= 2 + 2 = 4 sq.unit

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NCERT Solutions of class 12 maths Miscellaneous exercise -Application of Integrals

NCERT Solutions of class 12 maths Miscellaneous exercise -Application of Integrals

NCERT Solutions of Maths for Class 12 Miscellaneous exercise -Application of Integrals