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# NCERT Solutions of class 12 Maths exercise 8.3 -Application of Integrals

These NCERT Solutions of class 12 Maths exercise 8.3 of chapter 8-Application of Integrals are created by an expert CBSE teacher. The NCERT solutions written here are the solutions of class 12 NCERT maths exercise 8.3 of chapter 8-Application of Integrals. We are hundred percent sure that you will like and understand the concept of maths used in exercise 8.3-Application of Integrals because each solution is created by a step-by-step way, however, if you face any difficulty in understanding the method used here,you can write and suggest us in the comment box.

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In this exercise you will study the questions based on the enclosed areas by the curves and the lines.

Exercise 8.1-Application of Integration

Exercise 8.2-Application of Integration

Chapter 8-Miscellaneous Exercise Application of Iintegration

Q1.Find the area enclosed by the curve whose equations are : y = 2x², x =2 and x-axis

Ans.

The equation of the curve y = 2x², shows that it is passing through  the origin and symmetric about y axis.

The required area ABCD

$\boldsymbol{=\int_{2}^{3}y.dx}$

$\boldsymbol{=2\int_{2}^{3}x^{2}.dx}$

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

$\boldsymbol{=\frac{2}{3}\left [ 3^{3}-2^{3} \right ]}$

$\boldsymbol{=\frac{2}{3}\left ( 27-8 \right )}$

$\boldsymbol{=\frac{2}{3}\left ( 19 \right )}$

$\boldsymbol{=\frac{38}{3}\: sq.unit}$

Q2. Find the area enclosed by the curve whose equations are :y = 5x4 ,x=3.x=7, and x-axis

Ans.

The equation of curve(prabola) y = 5xshows that the curve is symmetric about y-axis and it is passing through the origin.

Area of ABCD

$\boldsymbol{=\int_{3}^{7}y.dx}$

$\boldsymbol{=\int_{3}^{7}5x^{4}.dx}$

$\boldsymbol{=5\int_{3}^{7}x^{4}.dx}$

$\boldsymbol{=5\left [ \frac{x^{5}}{5} \right ]_{3}^{7}}$

$\boldsymbol{=\left [ x^{5} \right ]_{3}^{7}}$

$\boldsymbol{=7^{5}-3^{5}=16807-243=16564\, unit^{2}}$

Q3.Find the area enclosed between the curve y²= 3x and line y = 6x.

Ans. We are given the curve y²= 3x and line y = 6x.

The equation of parabola shows that its vertex is at (0,0) and it is symmetric about the x-axis located at the right side of the y-axis

Solving both equations, we get point of intersections of line and parabola

(6x)² = 3x⇒ 36x² = 3x⇒ 12x = 1⇒ x = 1/12

y = 6 × 1/12 = 1/2

So, as the equation of line shows that it passes through the origin and we have got the point of intersection of the line and the curve is (1/12, 1/2)

Area covered by the curve and the line = Area enclosed by the curve (y²= 3x) OAB – area of the triangle OAB

The area enclosed by the curve (y²= 3x) OAB

$\boldsymbol{=\int_{0}^{1/12}y.dx}$

$\boldsymbol{=\int_{0}^{1/12}\sqrt{3x}.dx}$

$\boldsymbol{=\sqrt{3}\int_{0}^{1/12}\sqrt{x}.dx}$

$\boldsymbol{=\sqrt{3}\int_{0}^{1/12}x^{1/2}.dx=\sqrt{3}\left [ \frac{x^{3/2}}{3/2} \right ]_{0}^{1/12}}$

$\boldsymbol{=\frac{2\sqrt{3}}{3}\left ( \frac{1}{12} \right )^{3/2}}$

$\boldsymbol{=\frac{2\sqrt{3}}{3}.\frac{1}{24\sqrt{3}}}$

$\boldsymbol{=\frac{1}{36}sq.unit}$

Area of ΔOAB

$\boldsymbol{=\frac{1}{2}OA\times AB}$

$\boldsymbol{=\frac{1}{2}\times \frac{1}{12}\times \frac{1}{2}=\frac{1}{48}sq.unit}$

Area covered by the curve and the line

$\boldsymbol{=\frac{1}{36}-\frac{1}{48}=\frac{1}{144}sq.unit}$

Q4. Find the area enclosed by the curve y = 2x² and the lines y = 1 and y = 3 and the y-axis.

Ans. We are given the curve y = 2x² shows that its vertex is at (0,0) and is symmetric about the y-axis and the given lines are y =1 and y = 3

Solving the equation of curve for x

$\boldsymbol{x=\sqrt{\frac{y}{2}}}$

The area enclosed by the curve and the lines y = 1 and y=3 is ar FABDE

It is symmetric about y-axis, so

ar FABDE = 2× ar FABC

ar FABC

$\boldsymbol{=\frac{1}{\sqrt{2}}\int_{1}^{3}\sqrt{y}.dy}$

$\boldsymbol{=\frac{1}{\sqrt{2}}\int_{1}^{3}y^{1/2}.dy}$

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

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

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

ar FABDE

$\boldsymbol{=\frac{2\sqrt{2}}{3}\left ( 3\sqrt{3} -1\right )sq.unit}$

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Area covered by the curve and the line