Class 12 Maths Chapter 7 Integrals – Exercise 7.11 NCERT Solutions

Introduction

Exercise 7.11 focuses on applications of definite integrals in finding areas. Students learn how to calculate areas bounded by curves, coordinate axes, and lines using integration. This exercise is crucial for connecting calculus with geometry and practical applications.

Key Formulas

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

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

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

$$\text{Area}={\int }_a^b[f(x)-g(x)]dx\text{where }f(x)\ge g(x)$$

3. Area bounded by curve and y‑axis:

$$\text{Area}={\int }_c^{d}xdy$$

Common Mistakes

• Forgetting to take absolute value when curve lies below x‑axis.

• Incorrectly identifying limits of integration.

• Mixing up which curve is above the other.

• Arithmetic errors in evaluating definite integrals.

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

Q1. Find area under curve $y=x^2$ from $x=0$ to $x=3$.

$$\text{Area}={\int }_0^3{x}^{2}dx={\left[\frac{x^3}{3}\right]}_0^3=9$$

Q2. Find area under curve $y=2x$ from $x=0$ to $x=4$.

$$\text{Area}={\int }_0^{4}2xdx={\left[x^2\right]}_0^4=16$$

Q3. Find area under curve $y=\sin x$ from $x=0$ to $x=\pi$.

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

Q4. Find area under curve $y=\cos x$ from $x=0$ to $x=\frac{\pi }{2}$.

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

Q5. Find area between curves $y=x$ and $y=x^2$ from $x=0$ to $x=1$.

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

Q6. Find area between curves $y=\sin x$ and $y=\cos x$ from $x=0$ to $x=\frac{\pi }{4}$.

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

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

$$\text{Area}={\int }_0^1{e}^{x}dx={\left[e^x\right]}_0^1=e-1$$

Q8. Find area under curve $y=\mathrm{\mid }x\mathrm{\mid }$ from $x=-2$ to $x=2$.

$$\text{Area}={\int }_{-2}^0(-x)dx+{\int }_0^{2}xdx=2+2=4$$

Q9. Find area under curve $y=x^3$ from $x=0$ to $x=2$.

$$\text{Area}={\int }_0^2{x}^{3}dx={\left[\frac{x^4}{4}\right]}_0^2=4$$

Q10. Find area under curve $y=\ln (1+x)$ from $x=0$ to $x=1$.

$$\text{Area}={\int }_0^1\ln (1+x)dx={[x\ln (1+x)]}_0^1-{\int }_0^1\frac{x}{1+x}dx$$

$$=\ln 2-(1-\ln 2)=2\ln 2-1$$

FAQs (10)

FAQ1. What is definite integral used for? To calculate area under curves.

FAQ2. Why take absolute value in area? To ensure area is positive.

FAQ3. What is area between two curves? Difference of integrals of upper and lower curves.

FAQ4. What is unit of area? Square units.

FAQ5. Can integrals give negative values? Yes, but area is always taken positive.

FAQ6. Why is Exercise 7.11 important? It applies integration to geometry.

FAQ7. What is practical use of definite integrals? Used in physics, engineering, economics.

FAQ8. What is area under sine curve from 0 to $\pi$? It equals 2.

FAQ9. What is area under exponential curve from 0 to 1? It equals $e-1$.

FAQ10. What is area between parabola and line? Integral of difference of functions.

Conclusion

Exercise 7.11 covers applications of definite integrals in finding areas with solved examples and FAQs. Mastering these problems helps students connect calculus with geometry and real‑life applications.

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