Class 12 Maths Chapter 8 Application of Integrals – Exercise 8.4 NCERT Solutions

Introduction

Exercise 8.4 focuses on finding areas bounded by curves using definite integrals in advanced cases. Students learn how to calculate areas enclosed between curves and lines, often requiring symmetry, substitution, or splitting into multiple integrals. This exercise is essential for mastering applications of integrals in geometry, physics, and engineering.

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 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 parabola $y^2=4x$ and line $x=4$. Limits: $y=-4$ to $y=4$.

$$A={\int }_{-4}^4(4-\frac{y^2}{4})dy={[4y-\frac{y^3}{12}]}_{-4}^4=\frac{128}{3}$$

Q2. Find area bounded by circle $x^2+y^2=a^2$. Area of circle:

$$A=\pi a^2$$

Q3. Find area bounded by ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Area of ellipse:

$$A=\pi ab$$

Q4. Find area bounded by parabola $y^2=4ax$ and line $x=a$. Limits: $y=-2a$ to $y=2a$.

$$A={\int }_{-2a}^{2a}(a-\frac{y^2}{4a})dy=\frac{16a^2}{3}$$

Q5. Find area bounded by parabola $x^2=4ay$ and line $y=a$. Limits: $x=-2a$ to $x=2a$.

$$A={\int }_{-2a}^{2a}(a-\frac{x^2}{4a})dx=\frac{8a^2}{3}$$

Q6. Find area bounded by circle $x^2+y^2=r^2$ in first quadrant.

$$A=\frac{1}{4}\pi r^2$$

Q7. Find area bounded by ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in first quadrant.

$$A=\frac{1}{4}\pi ab$$

Q8. Find area bounded by parabola $y^2=4ax$ and x‑axis from $x=0$ to $x=a$.

$$A={\int }_0^a\sqrt{4ax}dx={\int }_0^{a}2\sqrt{ax}dx$$

$$=2\sqrt{a}{\int }_0^a\sqrt{x}dx=2\sqrt{a}\cdot \frac{2}{3}{a}^{3\mathrm{/}2}=\frac{4a^2}{3}$$

Q9. Find area bounded by parabola $x^2=4ay$ and y‑axis from $y=0$ to $y=a$.

$$A={\int }_0^a\sqrt{4ay}dy={\int }_0^{a}2\sqrt{ay}dy$$

$$=2\sqrt{a}{\int }_0^a\sqrt{y}dy=2\sqrt{a}\cdot \frac{2}{3}{a}^{3\mathrm{/}2}=\frac{4a^2}{3}$$

Q10. Find area bounded by circle $x^2+y^2=a^2$ and x‑axis. Half circle area:

$$A=\frac{1}{2}\pi a^2$$

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 of circle $x^2+y^2=a^2$? $\pi a^2$.

FAQ5. What is area of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$? $\pi ab$.

FAQ6. What is area bounded by parabola $y^2=4ax$ and line $x=a$? $\frac{16a^2}{3}$.

FAQ7. What is area bounded by parabola $x^2=4ay$ and line $y=a$? $\frac{8a^2}{3}$.

FAQ8. What is area of circle in first quadrant? $\frac{1}{4}\pi r^2$.

FAQ9. What is area of ellipse in first quadrant? $\frac{1}{4}\pi ab$.

FAQ10. Why is Exercise 8.4 important? It builds mastery of bounded areas using definite integrals in advanced cases.

Conclusion

Exercise 8.4 has 10 solved questions and 10 FAQs that strengthen your understanding of areas bounded by curves using definite integrals. This completes the Applications of Integrals chapter in Class 12 Maths.

visit:www.fuzymathacademy.com