Complete solutions for Class 12 Maths (NCERT) Exercise 7.1 on Integrals. Learn to find the anti-derivative of functions like $\sin 2x$, $\cos 3x$, and $(ax+b)^2$ using the method of inspection and applying fundamental integration formulas for polynomials and exponential functions.
This exercise introduces the basic concept of integration (finding the anti-derivative) using fundamental formulas and algebraic manipulation.
Table of Contents
Part 1: Anti-derivative by Inspection (1-5)
The goal is to find a function $G(x)$ such that $\frac{d}{dx} G(x) = f(x)$.
1. $\sin 2x$
We know $\frac{d}{dx}(\cos 2x) = -2 \sin 2x$.
Therefore, $\frac{d}{dx} \left(-\frac{1}{2} \cos 2x\right) = \sin 2x$.
Anti-derivative: $\mathbf{-\frac{1}{2} \cos 2x}$
2. $\cos 3x$
We know $\frac{d}{dx}(\sin 3x) = 3 \cos 3x$.
Therefore, $\frac{d}{dx} \left(\frac{1}{3} \sin 3x\right) = \cos 3x$.
Anti-derivative: $\mathbf{\frac{1}{3} \sin 3x}$
3. $e^{2x}$
We know $\frac{d}{dx}(e^{2x}) = 2 e^{2x}$.
Therefore, $\frac{d}{dx} \left(\frac{1}{2} e^{2x}\right) = e^{2x}$.
Anti-derivative: $\mathbf{\frac{1}{2} e^{2x}}$
4. $(ax + b)^2$
We know $\frac{d}{dx} (ax + b)^3 = 3(ax + b)^2 \cdot a$.
Therefore, $\frac{d}{dx} \left(\frac{1}{3a} (ax + b)^3\right) = (ax + b)^2$.
Anti-derivative: $\mathbf{\frac{1}{3a} (ax + b)^3}$
5. $\sin 2x – 4 e^{3x}$
We use the results from (1) and (3) adjusted for constants.
$$\frac{d}{dx} \left(-\frac{1}{2} \cos 2x\right) = \sin 2x$$
$$\frac{d}{dx} \left(C e^{3x}\right) = 3C e^{3x} \implies 4 e^{3x} \text{ requires } 3C = 4 \implies C = 4/3$$
$$\frac{d}{dx} \left(\frac{4}{3} e^{3x}\right) = 4 e^{3x}$$
Anti-derivative: $\mathbf{-\frac{1}{2} \cos 2x – \frac{4}{3} e^{3x}}$
Part 2: Integration using Formulas (6-20)
We use the formulas $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, $\int e^x dx = e^x + C$, and basic trigonometric integrals.
6. $\int (4e^{3x} + 1) dx$
$$\int (4e^{3x} + 1) dx = 4 \int e^{3x} dx + \int 1 dx$$
$$= 4 \left(\frac{e^{3x}}{3}\right) + x + C = \mathbf{\frac{4}{3} e^{3x} + x + C}$$
7. $\int x^2 \left(1 – \frac{1}{x^2}\right) dx$
First, simplify the integrand:
$$\int (x^2 – 1) dx = \int x^2 dx – \int 1 dx$$
$$= \mathbf{\frac{x^3}{3} – x + C}$$
8. $\int (ax^2 + bx + c) dx$
$$\int ax^2 dx + \int bx dx + \int c dx$$
$$= a \frac{x^3}{3} + b \frac{x^2}{2} + c x + C = \mathbf{\frac{ax^3}{3} + \frac{bx^2}{2} + cx + C}$$
9. $\int (2x^2 + e^x) dx$
$$\int 2x^2 dx + \int e^x dx$$
$$= 2 \frac{x^3}{3} + e^x + C = \mathbf{\frac{2x^3}{3} + e^x + C}$$
10. $\int \left(\sqrt{x} – \frac{1}{\sqrt{x}}\right)^2 dx$
First, expand the square:
$$\int \left(x + \frac{1}{x} – 2\sqrt{x} \cdot \frac{1}{\sqrt{x}}\right) dx = \int \left(x + \frac{1}{x} – 2\right) dx$$
$$= \int x dx + \int \frac{1}{x} dx – \int 2 dx$$
$$= \mathbf{\frac{x^2}{2} + \log|x| – 2x + C}$$
11. $\int \frac{x^3 + 5x^2 – 4}{x^2} dx$
Separate the terms by dividing by $x^2$:
$$\int \left(\frac{x^3}{x^2} + \frac{5x^2}{x^2} – \frac{4}{x^2}\right) dx = \int \left(x + 5 – 4x^{-2}\right) dx$$
$$= \frac{x^2}{2} + 5x – 4 \frac{x^{-2+1}}{-2+1} + C = \frac{x^2}{2} + 5x – 4 \frac{x^{-1}}{-1} + C$$
$$= \mathbf{\frac{x^2}{2} + 5x + \frac{4}{x} + C}$$
12. $\int \frac{x^3 + 3x + 4}{\sqrt{x}} dx$
Write $\sqrt{x}$ as $x^{1/2}$ and separate the terms:
$$\int \left(\frac{x^3}{x^{1/2}} + \frac{3x}{x^{1/2}} + \frac{4}{x^{1/2}}\right) dx = \int \left(x^{5/2} + 3x^{1/2} + 4x^{-1/2}\right) dx$$
$$= \frac{x^{7/2}}{7/2} + 3 \frac{x^{3/2}}{3/2} + 4 \frac{x^{1/2}}{1/2} + C$$
$$= \mathbf{\frac{2}{7} x^{7/2} + 2x^{3/2} + 8x^{1/2} + C}$$
13. $\int \frac{x^3 – x^2 + x – 1}{x – 1} dx$
Factor the numerator:
$$\int \frac{x^2(x – 1) + 1(x – 1)}{x – 1} dx = \int \frac{(x – 1)(x^2 + 1)}{x – 1} dx$$
For $x \ne 1$, this simplifies to:
$$\int (x^2 + 1) dx = \frac{x^3}{3} + x + C$$
$$= \mathbf{\frac{x^3}{3} + x + C}$$
14. $\int (1 – x) \sqrt{x} dx$
Simplify the integrand:
$$\int (\sqrt{x} – x \sqrt{x}) dx = \int (x^{1/2} – x^{3/2}) dx$$
$$= \frac{x^{3/2}}{3/2} – \frac{x^{5/2}}{5/2} + C$$
$$= \mathbf{\frac{2}{3} x^{3/2} – \frac{2}{5} x^{5/2} + C}$$
15. $\int \sqrt{x} (3x^2 + 2x + 3) dx$
Write $\sqrt{x}$ as $x^{1/2}$ and distribute:
$$\int (3x^{2 + 1/2} + 2x^{1 + 1/2} + 3x^{1/2}) dx = \int (3x^{5/2} + 2x^{3/2} + 3x^{1/2}) dx$$
$$= 3 \frac{x^{7/2}}{7/2} + 2 \frac{x^{5/2}}{5/2} + 3 \frac{x^{3/2}}{3/2} + C$$
$$= \mathbf{\frac{6}{7} x^{7/2} + \frac{4}{5} x^{5/2} + 2x^{3/2} + C}$$
16. $\int (2x – 3 \cos x + e^x) dx$
$$\int 2x dx – \int 3 \cos x dx + \int e^x dx$$
$$= 2 \frac{x^2}{2} – 3 \sin x + e^x + C$$
$$= \mathbf{x^2 – 3 \sin x + e^x + C}$$
17. $\int (2x^2 – 3 \sin x + 5 \sqrt{x}) dx$
$$\int 2x^2 dx – \int 3 \sin x dx + \int 5 x^{1/2} dx$$
$$= 2 \frac{x^3}{3} – 3 (-\cos x) + 5 \frac{x^{3/2}}{3/2} + C$$
$$= \mathbf{\frac{2x^3}{3} + 3 \cos x + \frac{10}{3} x^{3/2} + C}$$
18. $\int \sec x (\sec x + \tan x) dx$
Distribute $\sec x$:
$$\int (\sec^2 x + \sec x \tan x) dx$$
$$= \int \sec^2 x dx + \int \sec x \tan x dx$$
$$= \mathbf{\tan x + \sec x + C}$$
19. $\int \frac{\sec^2 x}{\csc^2 x} dx$
Use trigonometric identities $\sec^2 x = 1/\cos^2 x$ and $\csc^2 x = 1/\sin^2 x$:
$$\int \frac{1/\cos^2 x}{1/\sin^2 x} dx = \int \frac{\sin^2 x}{\cos^2 x} dx = \int \tan^2 x dx$$
Use the identity $\tan^2 x = \sec^2 x – 1$:
$$\int (\sec^2 x – 1) dx = \int \sec^2 x dx – \int 1 dx$$
$$= \mathbf{\tan x – x + C}$$
20. $\int \frac{2 – 3 \sin x}{\cos^2 x} dx$
Separate the fraction:
$$\int \left(\frac{2}{\cos^2 x} – \frac{3 \sin x}{\cos^2 x}\right) dx$$
Use identities: $\frac{1}{\cos^2 x} = \sec^2 x$ and $\frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x$:
$$\int (2 \sec^2 x – 3 \sec x \tan x) dx$$
$$= 2 \int \sec^2 x dx – 3 \int \sec x \tan x dx$$
$$= \mathbf{2 \tan x – 3 \sec x + C}$$
Part 3: Multiple Choice Questions (21-22)
21. The anti derivative of $\sqrt{x} + \frac{1}{\sqrt{x}}$ equals:
$$\int \left(x^{1/2} + x^{-1/2}\right) dx = \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C$$
$$= \frac{2}{3} x^{3/2} + 2 x^{1/2} + C$$
Write $x^{3/2}$ as $x \sqrt{x}$ and $x^{1/2}$ as $\sqrt{x}$:
$$= \frac{2}{3} x \sqrt{x} + 2 \sqrt{x} + C$$
This matches option (C). (Note: The fractional exponents in the options use standard notation, e.g., $x^{3/2}$ is written as $x^{3/2}$)
$$\text{(C) } \mathbf{\frac{2}{3} x^{3/2} + 2 x^{1/2} + C}$$
22. If $\frac{d}{dx} f(x) = 4x^3 – \frac{3}{x^4}$ such that $f(2) = 0$. Then $f(x)$ is:
- Find $f(x)$ by integrating the derivative:$$f(x) = \int \left(4x^3 – 3x^{-4}\right) dx = 4 \frac{x^4}{4} – 3 \frac{x^{-3}}{-3} + C$$$$f(x) = x^4 + x^{-3} + C = x^4 + \frac{1}{x^3} + C$$
- Use the condition $f(2) = 0$ to find $C$:$$f(2) = (2)^4 + \frac{1}{(2)^3} + C = 0$$$$16 + \frac{1}{8} + C = 0 \implies \frac{128 + 1}{8} + C = 0$$$$C = -\frac{129}{8}$$
- Final $f(x)$:$$f(x) = x^4 + \frac{1}{x^3} – \frac{129}{8}$$
This matches option (A).
$$\text{(A) } \mathbf{x^4 + \frac{1}{x^3} – \frac{129}{8}}$$