Class 12 Maths Chapter 7 Integrals – Exercise 7.9 NCERT Solutions

Introduction

Exercise 7.9 focuses on definite integrals and their properties. Students learn how to apply properties of definite integrals to simplify calculations and solve problems efficiently. This exercise is important for mastering advanced integration techniques and preparing for competitive exams.

Key Properties of Definite Integrals

1. Property 1:

$${\int }_a^{b}f(x)dx={\int }_a^{b}f(t)dt$$

2. Property 2:

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

3. Property 3:

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

4. Property 4:

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

Common Mistakes

• Forgetting to apply symmetry property correctly.

• Confusing limits when reversing integration.

• Arithmetic errors in splitting integrals.

• Misidentifying the function transformation in property 4.

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

Q1. Evaluate ${\int }_0^1(x^2+(1-x{)}^2)dx$. Using property 4:

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

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

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

Q2. Evaluate ${\int }_0^{\pi }{\sin }^{2}xdx$.

$${\sin }^{2}x=\frac{1-\cos 2x}{2}$$

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

Q3. Evaluate ${\int }_0^{\pi }{\cos }^{2}xdx$.

$${\cos }^{2}x=\frac{1+\cos 2x}{2}$$

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

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

$${\int }_0^1\frac{1}{1+x}dx={[\ln (1+x)]}_0^1=\ln 2$$

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

$$\frac{x}{1+x}=1-\frac{1}{1+x}$$

$${\int }_0^1\frac{x}{1+x}dx={\int }_0^{1}1dx-{\int }_0^1\frac{1}{1+x}dx=1-\ln 2$$

Q6. Evaluate ${\int }_0^1\ln (1+x)dx$. Integration by parts:

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

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

Q7. Evaluate ${\int }_0^{\pi }x\sin xdx$. Integration by parts:

$$={[-x\cos x]}_0^{\pi }+{\int }_0^{\pi }\cos xdx=\pi +0=\pi$$

Q8. Evaluate ${\int }_0^{\pi }x\cos xdx$. Integration by parts:

$$={[x\sin x]}_0^{\pi }-{\int }_0^{\pi }\sin xdx=0-2=-2$$

Q9. Evaluate ${\int }_0^1(x^2+(1-x{)}^2)dx$. Already solved in Q1 = $\frac{2}{3}$.

Q10. Evaluate ${\int }_0^{\pi }\sin x\cos xdx$.

$$\sin x\cos x=\frac{1}{2}\sin 2x$$

$${\int }_0^{\pi }\sin x\cos xdx=\frac{1}{2}{\int }_0^{\pi }\sin 2xdx=0$$

FAQs (10)

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

FAQ2. What is property of symmetry in integrals? ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$.

FAQ3. Why use properties of integrals? To simplify calculations.

FAQ4. What is unit of definite integral? Depends on function; often square units for area.

FAQ5. Can definite integrals be negative? Yes, but area is taken positive.

FAQ6. What is practical use of definite integrals? Used in physics, engineering, economics.

FAQ7. What is property of reversing limits? ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$.

FAQ8. Why split integrals? To simplify evaluation over sub‑intervals.

FAQ9. What is median property of integrals? Symmetry property helps in simplification.

FAQ10. Why is Exercise 7.9 important? It builds mastery of definite integrals and properties.

Conclusion

Exercise 7.9 covers properties of definite integrals with solved examples and FAQs. Mastering these problems helps students simplify integration and apply it effectively in advanced mathematics.

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