Class 12 Maths Chapter 8 Application of Integrals – Exercise 8.3 NCERT Solutions

Introduction

Exercise 8.3 focuses on finding areas bounded by curves and lines using definite integrals in more complex cases. Students learn how to calculate areas enclosed between curves, coordinate axes, and intersecting lines. This exercise is essential for mastering applications of integrals in geometry and real‑life contexts.

Formulas Used

1. Area between two curves $y=f(x)$ and $y=g(x)$:

$$A={\int }_a^b[f(x)-g(x)]dx,f(x)\ge g(x)$$

2. Area bounded by curve and x‑axis:

$$A={\int }_a^b\mathrm{\mid }f(x)\mathrm{\mid }dx$$

3. Area bounded by curve and y‑axis:

$$A={\int }_c^{d}f(y)dy$$

4. Symmetry Property: If curve is symmetric, calculate area for half and double it.

Students Frequently Make Mistakes

• Forgetting to find intersection points of curves correctly.

• Errors in setting integration limits.

• Confusing which curve is upper/lower in given interval.

• Ignoring absolute value when curve lies below x‑axis.

• Skipping symmetry to simplify calculations.

NCERT Questions with Step‑by‑Step Solutions (10)

Q1. Find area bounded by $y=x^2$ and $y=\mathrm{\mid }x\mathrm{\mid }$. Intersection at $x=-1,0,1$. Area = 2 × ${\int }_0^1(x-x^2)dx$.

$$=2{[\frac{x^2}{2}-\frac{x^3}{3}]}_0^1=\frac{1}{3}$$

Q2. Find area bounded by $y=x^2$ and $y=2x$. Intersection at $x=0,2$.

$$A={\int }_0^2(2x-x^2)dx={[x^2-\frac{x^3}{3}]}_0^2=\frac{4}{3}$$

Q3. Find area bounded by $y=\sin x$ and $y=\cos x$ between $x=0$ and $x=\pi \mathrm{/}4$.

$$A={\int }_0^{\pi \mathrm{/}4}(\cos x-\sin x)dx=[\sin x+\cos x{]}_0^{\pi \mathrm{/}4}=(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2})-(0+1)=\sqrt{2}-1$$

Q4. Find area bounded by $y=\sin x$ and $y=\cos x$ between $x=\pi \mathrm{/}4$ and $x=\pi \mathrm{/}2$.

$$A={\int }_{\pi \mathrm{/}4}^{\pi \mathrm{/}2}(\sin x-\cos x)dx=[-\cos x-\sin x{]}_{\pi \mathrm{/}4}^{\pi \mathrm{/}2}=(0-1)-(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2})=1-\sqrt{2}$$

Area = $\sqrt{2}-1$ (positive).

Q5. Find area bounded by $y=x^2$ and $y=4$. Intersection at $x=-2,2$.

$$A={\int }_{-2}^2(4-x^2)dx={[4x-\frac{x^3}{3}]}_{-2}^2=\frac{32}{3}$$

Q6. Find area bounded by $y=\sqrt{x}$ and $y=x^2$ between $x=0$ and $x=1$.

$$A={\int }_0^1(\sqrt{x}-x^2)dx={[\frac{2}{3}{x}^{3\mathrm{/}2}-\frac{x^3}{3}]}_0^1=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$$

Q7. Find area bounded by $y=\sin x$ and x‑axis from $x=0$ to $x=\pi$.

$$A={\int }_0^{\pi }\sin xdx=[-\cos x{]}_0^{\pi }=2$$

Q8. Find area bounded by $y=\cos x$ and x‑axis from $x=0$ to $x=\pi \mathrm{/}2$.

$$A={\int }_0^{\pi \mathrm{/}2}\cos xdx=[\sin x{]}_0^{\pi \mathrm{/}2}=1$$

Q9. Find area bounded by $y=\tan x$ and x‑axis from $x=0$ to $x=\pi \mathrm{/}4$.

$$A={\int }_0^{\pi \mathrm{/}4}\tan xdx=[-\ln \mathrm{\mid }\cos x\mathrm{\mid }{]}_0^{\pi \mathrm{/}4}=\ln \sqrt{2}$$

Q10. Find area bounded by $y={\sec }^{2}x$ and x‑axis from $x=0$ to $x=\pi \mathrm{/}4$.

$$A={\int }_0^{\pi \mathrm{/}4}{\sec }^{2}xdx=[\tan x{]}_0^{\pi \mathrm{/}4}=1$$

FAQs (10)

FAQ1. What is formula for area between two curves? ${\int }_a^b[f(x)-g(x)]dx$.

FAQ2. Why find intersection points first? They define limits of integration.

FAQ3. What if curve lies below x‑axis? Take absolute value of integral.

FAQ4. What is area between $y=x^2$ and $y=2x$? $\frac{4}{3}$.

FAQ5. What is area between $y=\sin x$ and $y=\cos x$ from 0 to $\pi \mathrm{/}4$? $\sqrt{2}-1$.

FAQ6. What is area under $y=\sin x$ from 0 to $\pi$?

FAQ7. What is area under $y=\cos x$ from 0 to $\pi \mathrm{/}2$?

FAQ8. What is area between $y=\sqrt{x}$ and $y=x^2$ from 0 to 1? $\frac{1}{3}$.

FAQ9. Why use symmetry in integrals? It simplifies calculations.

FAQ10. Why is Exercise 8.3 important? It builds mastery of bounded areas using definite integrals.

Conclusion

Exercise 8.3 has 10 solved questions and 10 FAQs that strengthen your understanding of areas bounded by curves and lines using definite integrals. This builds the foundation for advanced applications of integrals in Class 12 Maths.

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