Class 12 Maths Chapter 8 Application of Integrals – Exercise 8.1 NCERT Solutions

Introduction

Exercise 8.1 introduces the application of definite integrals in finding areas under curves. Students learn how to calculate the area bounded by curves, coordinate axes, and lines using integration. This exercise is fundamental for geometry, physics, and probability applications.

Formulas Used

1. Area under curve $y=f(x)$ from $x=a$ to $x=b$:

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

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

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

3. Area under curve $x=f(y)$ from $y=c$ to $y=d$:

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

Students Frequently Make Mistakes

• Forgetting to identify upper and lower curves correctly.

• Errors in setting limits of integration.

• Confusing area with signed integral (negative values).

• Skipping modulus when area must be positive.

• Misinterpreting symmetry in curves.

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

Q1. Find area under $y=x$ from $x=0$ to $x=2$.

$$A={\int }_0^{2}xdx={\left[\frac{x^2}{2}\right]}_0^2=2$$

Q2. Find area under $y=x^2$ from $x=0$ to $x=1$.

$$A={\int }_0^1{x}^{2}dx={\left[\frac{x^3}{3}\right]}_0^1=\frac{1}{3}$$

Q3. Find area under $y=\sqrt{x}$ from $x=0$ to $x=4$.

$$A={\int }_0^4\sqrt{x}dx={\left[\frac{2}{3}{x}^{3\mathrm{/}2}\right]}_0^4=\frac{16}{3}$$

Q4. Find area under $y=\sin x$ from $x=0$ to $x=\pi$.

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

Q5. Find area under $y=\cos x$ from $x=0$ to $x=\pi \mathrm{/}2$.

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

Q6. Find area between curves $y=x$ and $y=x^2$ from $x=0$ to $x=1$.

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

Q7. Find area under $y=e^x$ from $x=0$ to $x=1$.

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

Q8. Find area under $y=\ln x$ from $x=1$ to $x=e$.

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

Q9. Find area under $y=\tan x$ 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 under $y={\sec }^{2}x$ 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 under curve? ${\int }_a^{b}f(x)dx$.

FAQ2. What is formula for area between two curves? ${\int }_a^b[f(x)-g(x)]dx$.

FAQ3. Why area always positive? Because it represents physical region.

FAQ4. What is area under $y=x$ from 0 to 2?

FAQ5. What is area under $y=x^2$ from 0 to 1? $\frac{1}{3}$.

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 $y=x$ and $y=x^2$ from 0 to 1? $\frac{1}{6}$.

FAQ9. Why use definite integrals for area? They give exact bounded region measure.

FAQ10. Why is Exercise 8.1 important? It builds foundation for applications of integrals in geometry and physics.

Conclusion

Exercise 8.1 has 10 solved questions and 10 FAQs that strengthen your understanding of areas under curves using definite integrals. This begins the Applications of Integrals chapter in Class 12 Maths.

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