Class 12 Maths Chapter 7 Integrals – Exercise 7.8 NCERT Solutions

Introduction

Exercise 7.8 focuses on definite integrals as areas under curves. Students learn how to apply integration to calculate areas bounded by curves, coordinate axes, and lines. This exercise is crucial for connecting integration with geometry and real‑life applications.

Key Formulas

1. Definite Integral as Area:

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

2. Area between Curve and x‑axis:

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

3. Area between Two Curves:

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

where $f(x)\ge g(x)$ in $[a,b]$.

Common Mistakes

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

• Incorrectly identifying limits of integration.

• Mixing up which curve is on top when finding area between two curves.

• 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 $x=2$.

$$\text{Area}={\int }_0^2{x}^{2}dx={\left[\frac{x^3}{3}\right]}_0^2=\frac{8}{3}$$

Q2. Find area under curve $y=2x$ from $x=0$ to $x=3$.

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

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 under curve $y=x$ from $x=0$ to $x=4$.

$$\text{Area}={\int }_0^{4}xdx={\left[\frac{x^2}{2}\right]}_0^4=8$$

Q6. Find area under curve $y=x^3$ from $x=0$ to $x=2$.

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

Q7. Find area under curve $y=\mathrm{\mid }x\mathrm{\mid }$ from $x=-2$ to $x=2$.

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

Q8. Find area between curves $y=x^2$ and $y=x$ from $x=0$ to $x=1$.

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

Q9. Find area between curves $y=\sin x$ and $y=\cos x$ from $x=0$ to $x=\frac{\pi }{4}$.

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

Q10. Find area under curve $y=e^x$ from $x=0$ to $x=1$.

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

FAQs (10)

FAQ1. What is definite integral? It represents area under curve between given limits.

FAQ2. Why use absolute value in area? To ensure area is positive even if curve lies below x‑axis.

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 7.8 important? It connects integration with 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 7.8 covers definite integrals as areas under curves with solved examples and FAQs. Mastering these problems helps students apply integration to geometry and real‑life applications.

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