Class 12 Maths Chapter 8 Application of Integrals – Exercise 8.2 NCERT Solutions

Introduction

Exercise 8.2 focuses on finding areas bounded by curves and lines using definite integrals. Students learn how to calculate the area enclosed between two curves, the coordinate axes, and given limits. This exercise is essential for geometry, calculus, and real‑life applications such as physics and economics.

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:

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

Students Frequently Make Mistakes

• Forgetting to identify which curve is above the other.

• Errors in setting correct limits of integration.

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

• Skipping symmetry property to simplify calculations.

• Misinterpreting geometric meaning of definite integral.

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

Q1. Find area bounded by $y=x^2$ and $y=x$ between $x=0$ and $x=1$.

$$A={\int }_0^1(x-x^2)dx={[\frac{x^2}{2}-\frac{x^3}{3}]}_0^1=\frac{1}{6}$$

Q2. Find area bounded by $y=x^2$ and $y=4$ between $x=-2$ and $x=2$.

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

Q3. Find area bounded by $y=\sin x$ and x‑axis from $x=0$ to $x=\pi$.

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

Q4. Find area bounded by $y=\cos x$ and x‑axis from $x=0$ to $x=\pi \mathrm{/}2$.

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

Q5. Find area bounded by $y=e^x$ and x‑axis from $x=0$ to $x=1$.

$$A={\int }_0^1{e}^{x}dx=[e^x{]}_0^1=e-1$$

Q6. Find area bounded by $y=\ln x$ and x‑axis from $x=1$ to $x=e$.

$$A={\int }_1^e\ln xdx=[x\ln x-x{]}_1^e=(e-1)$$

Q7. Find area bounded by $y=\tan x$ and x‑axis from $x=0$ to $x=\pi \mathrm{/}4$.

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

Q8. Find area bounded by $y={\sec }^{2}x$ and x‑axis from $x=0$ to $x=\pi \mathrm{/}4$.

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

Q9. Find area bounded by $y=x^2$ and $y=2x$ between $x=0$ and $x=2$.

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

Q10. Find area bounded by $y=x^3$ and $y=x$ between $x=0$ and $x=1$.

$$A={\int }_0^1(x-x^3)dx={[\frac{x^2}{2}-\frac{x^4}{4}]}_0^1=\frac{1}{4}$$

FAQs (10)

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

FAQ2. Why use absolute value in area? Because area is always positive.

FAQ3. What is area between $y=x$ and $y=x^2$ from 0 to 1? $\frac{1}{6}$.

FAQ4. What is area under $y=\sin x$ from 0 to $\pi$?

FAQ5. What is area under $y=\cos x$ from 0 to $\pi \mathrm{/}2$?

FAQ6. What is area under $y=e^x$ from 0 to 1? $e-1$.

FAQ7. What is area under $y=\ln x$ from 1 to $e$? $e-1$.

FAQ8. What is area under $y=\tan x$ from 0 to $\pi \mathrm{/}4$? $\ln \sqrt{2}$.

FAQ9. What is area between $y=x^2$ and $y=2x$ from 0 to 2? $\frac{4}{3}$.

FAQ10. Why is Exercise 8.2 important? It builds foundation for calculating bounded areas using definite integrals.

Conclusion

Exercise 8.2 has 10 solved questions and 10 FAQs that strengthen your understanding of areas bounded by curves using definite integrals. This builds the foundation for advanced applications of integrals in Class 12 Maths.

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