Class 12 Maths Chapter 7 Integrals – Exercise 7.6 NCERT Solutions

Introduction

Exercise 7.6 focuses on properties of definite integrals. Students learn how to simplify definite integrals using symmetry, periodicity, and transformation rules. This exercise is essential for solving integrals efficiently and reducing complex problems into standard forms.

Properties Used

1. Reversal of Limits:

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

2. Zero Interval:

$${\int }_a^{a}f(x)dx=0$$

3. Additivity:

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

4. Symmetry Property:

$${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$$

5. Even/Odd Functions:

   • If $f(x)$ is even:

$${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$$

• If $f(x)$ is odd:

$${\int }_{-a}^{a}f(x)dx=0$$

Students Frequently Make Mistakes

• Forgetting to apply symmetry property correctly.

• Confusing even and odd functions.

• Errors in reversing limits.

• Skipping modulus in logarithmic evaluations.

• Not splitting integrals properly at intermediate points.

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

Q1. Evaluate ${\int }_0^{a}f(x)dx$ using property $f(x)=f(a-x)$.

$${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$$

Q2. Evaluate ${\int }_{-a}^a{x}^{2}dx$. Even function ⇒

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

Q3. Evaluate ${\int }_{-a}^a{x}^{3}dx$. Odd function ⇒ integral = 0.

Q4. Evaluate ${\int }_0^{\pi }\sin xdx$.

$$[-\cos x{]}_0^{\pi }=2$$

Q5. Evaluate ${\int }_0^{\pi }\cos xdx$.

$$[\sin x{]}_0^{\pi }=0$$

Q6. Evaluate ${\int }_0^a(x+a-x)dx$.

$${\int }_0^{a}adx=a^2$$

Q7. Evaluate ${\int }_0^1(x^2+(1-x{)}^2)dx$.

$${\int }_0^1(x^2+1-2x+x^2)dx={\int }_0^1(2x^2-2x+1)dx$$

$$={[\frac{2x^3}{3}-x^2+x]}_0^1=\frac{2}{3}$$

Q8. Evaluate ${\int }_0^{a}x(a-x)dx$.

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

Q9. Evaluate ${\int }_0^{\pi }{\sin }^{2}xdx$. Use identity: ${\sin }^{2}x=\frac{1-\cos 2x}{2}$.

$${\int }_0^{\pi }{\sin }^{2}xdx=\frac{1}{2}{\int }_0^{\pi }(1-\cos 2x)dx$$

$$=\frac{1}{2}{[x-\frac{\sin 2x}{2}]}_0^{\pi }=\frac{\pi }{2}$$

Q10. Evaluate ${\int }_0^{\pi }{\cos }^{2}xdx$. Use identity: ${\cos }^{2}x=\frac{1+\cos 2x}{2}$.

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

$$=\frac{1}{2}{[x+\frac{\sin 2x}{2}]}_0^{\pi }=\frac{\pi }{2}$$

FAQs (10)

FAQ1. What is property of reversal of limits? ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$.

FAQ2. What is property of symmetry? ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$.

FAQ3. What is integral of even function over $[-a,a]$? Twice integral from 0 to $a$.

FAQ4. What is integral of odd function over $[-a,a]$? Zero.

FAQ5. Why split integrals? To simplify using additivity property.

FAQ6. What is ${\int }_a^{a}f(x)dx$? Zero.

FAQ7. Why use symmetry in definite integrals? It reduces complexity.

FAQ8. What is integral of ${\sin }^{2}x$ over $[0,\pi ]$? $\frac{\pi }{2}$.

FAQ9. What is integral of ${\cos }^{2}x$ over $[0,\pi ]$? $\frac{\pi }{2}$.

FAQ10. Why is Exercise 7.6 important? It builds foundation for properties of definite integrals and efficient problem solving.

Conclusion

Exercise 7.6 has 10 solved questions and 10 FAQs that strengthen your understanding of properties of definite integrals. This builds the foundation for advanced calculus applications in Class 12 Maths.

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