Class 12 Maths Chapter 7 Integrals – Exercise 7.10 NCERT Solutions

Introduction

Exercise 7.10 focuses on definite integrals and applications of properties. Students practice evaluating integrals using symmetry, periodicity, and transformations. This exercise strengthens problem‑solving skills and prepares students for advanced calculus and competitive exams.

Key Properties Recap

1. Symmetry Property:

$${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$$

2. Reversal of Limits:

$${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$$

3. Splitting Interval:

$${\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx$$

Common Mistakes

• Forgetting to apply symmetry correctly.

• Confusing periodic functions with symmetric ones.

• Arithmetic errors in trigonometric integrals.

• Not checking which function is dominant in area problems.

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

Q1. Evaluate ${\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx$. Using property:

$$I={\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}dx$$

Q2. Evaluate ${\int }_0^{\pi }\frac{x}{1+{\sin }^{2}x}dx$. By symmetry:

$$I=\frac{\pi }{2}{\int }_0^{\pi }\frac{1}{1+{\sin }^{2}x}dx$$

Q3. Evaluate ${\int }_0^{\pi }\frac{x\cos x}{1+{\sin }^{2}x}dx$. Using property:

$$I=\frac{\pi }{2}{\int }_0^{\pi }\frac{\cos x}{1+{\sin }^{2}x}dx$$

Q4. Evaluate ${\int }_0^{\pi }\frac{x}{1+{\cos }^{2}x}dx$.

$$I=\frac{\pi }{2}{\int }_0^{\pi }\frac{1}{1+{\cos }^{2}x}dx$$

Q5. Evaluate ${\int }_0^{\pi }\frac{x\sin x}{1+{\sin }^{2}x}dx$.

$$I=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin x}{1+{\sin }^{2}x}dx$$

Q6. Evaluate ${\int }_0^{\pi }\frac{x}{1+{\tan }^{2}x}dx$. Since $1+{\tan }^{2}x={\sec }^{2}x$:

$$I={\int }_0^{\pi }x{\cos }^{2}xdx$$

Q7. Evaluate ${\int }_0^{\pi }\frac{x}{1+{\cot }^{2}x}dx$. Since $1+{\cot }^{2}x={\csc }^{2}x$:

$$I={\int }_0^{\pi }x{\sin }^{2}xdx$$

Q8. Evaluate ${\int }_0^{\pi }\frac{x}{1+\cos x}dx$. By property:

$$I=\frac{\pi }{2}{\int }_0^{\pi }\frac{1}{1+\cos x}dx$$

Q9. Evaluate ${\int }_0^{\pi }\frac{x}{1+\sin x}dx$.

$$I=\frac{\pi }{2}{\int }_0^{\pi }\frac{1}{1+\sin x}dx$$

Q10. Evaluate ${\int }_0^{\pi }\frac{x}{1+{\cos }^{2}x}dx$.

$$I=\frac{\pi }{2}{\int }_0^{\pi }\frac{1}{1+{\cos }^{2}x}dx$$

FAQs (10)

FAQ1. What is symmetry property of integrals? ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$.

FAQ2. Why use symmetry in definite integrals? It simplifies evaluation.

FAQ3. What is periodicity property? Integrals of periodic functions repeat over intervals.

FAQ4. What is reversal of limits? ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$.

FAQ5. Why split integrals? To evaluate over manageable intervals.

FAQ6. Can definite integrals be negative? Yes, depending on function values.

FAQ7. What is unit of definite integral? Depends on function; often square units for area.

FAQ8. What is practical use of definite integrals? Used in physics, engineering, economics.

FAQ9. Why is Exercise 7.10 important? It teaches advanced properties of integrals.

FAQ10. What is difference between mean value and definite integral? Mean value is average; definite integral is total accumulation.

Conclusion

Exercise 7.10 covers applications of properties of definite integrals with solved examples and FAQs. Mastering these problems helps students simplify complex integrals and apply calculus effectively in advanced mathematics.

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