Class 12th Mathematics Part II CBSE Solution - Exercise 7.1
Find an anti-derivative (or integral) of the following functions by the method of inspection.
Function: \(\sin 2x\)
Answer:
Method: To find the anti derivative of a function by inspection
Steps:
- In this method we look for a function whose derivative is the given function. For Example: if we need to find anti derivative of \(x^2\), we know that derivative of \(x^3\) is \(3x^2\). Therefore, the variable terms comes out to be the same.
- After that balance out the coefficients of variables by dividing and multiplying suitable terms. From above example if we divide \(x^3\) by 3 we will get the answer as \(x^2\). Hence, we can say that anti derivative of \(x^2\) is \(x^3/3\).
Now, similarly,
We know that:
\(\Rightarrow \sin 2x = -\frac{1}{2} \frac{d}{dx}(\cos 2x)\)
\(\Rightarrow \sin 2x = \frac{d}{dx}(-\frac{1}{2} \cos 2x)\)
Therefore, the anti-derivative of \(\sin 2x\) is \(-\frac{1}{2} \cos 2x\).
Find an anti-derivative (or integral) of the following functions by the method of inspection.
Function: \(\cos 3x\)
Answer:
We know that:
\(\Rightarrow \cos 3x = \frac{1}{3} \frac{d}{dx}(\sin 3x)\)
\(\Rightarrow \cos 3x = \frac{d}{dx}(\frac{1}{3} \sin 3x)\)
Therefore, the anti-derivative of \(\cos 3x\) is \(\frac{1}{3} \sin 3x\).
Find an anti-derivative (or integral) of the following functions by the method of inspection.
Function: \(e^{2x}\)
Answer:
We know that:
\(\Rightarrow e^{2x} = \frac{1}{2} \frac{d}{dx}(e^{2x})\)
\(\Rightarrow e^{2x} = \frac{d}{dx}(\frac{1}{2} e^{2x})\)
Therefore, the anti-derivative of \(e^{2x}\) is \(\frac{1}{2} e^{2x}\).
Find an anti-derivative (or integral) of the following functions by the method of inspection.
Function: \((ax + b)^2\)
Answer:
We know that:
\(\Rightarrow (ax + b)^2 = \frac{1}{3a} \frac{d}{dx}(ax + b)^3\)
\(\Rightarrow (ax + b)^2 = \frac{d}{dx}\left( \frac{1}{3a}(ax + b)^3 \right)\)
Therefore, the anti-derivative of \((ax + b)^2\) is \(\frac{1}{3a}(ax + b)^3\).
Find an anti-derivative (or integral) of the following functions by the method of inspection.
Function: \(\sin 2x - 4 e^{3x}\)
Answer:
We know that:
\(\Rightarrow -\frac{1}{2} \frac{d}{dx}(\cos 2x) = \sin 2x\)
\(\Rightarrow \frac{d}{dx}\left(-\frac{1}{2} \cos 2x\right) = \sin 2x\)
Therefore, the anti-derivative of \(\sin 2x\) is \(-\frac{1}{2} \cos 2x\) ...(1)
Also,
\(\Rightarrow \frac{4}{3} \frac{d}{dx}(e^{3x}) = 4e^{3x}\)
\(\Rightarrow \frac{d}{dx}\left(\frac{4}{3} e^{3x}\right) = 4e^{3x}\)
Therefore, the anti-derivative of \(4e^{3x}\) is \(\frac{4}{3} e^{3x}\) ...(2)
From (1) and (2), we get,
\(\frac{d}{dx}\left(-\frac{1}{2} \cos 2x - \frac{4}{3} e^{3x}\right) = \sin 2x - 4e^{3x}\)
Therefore, the anti-derivative of \(\sin 2x - 4 e^{3x}\) is \(-\frac{1}{2} \cos 2x - \frac{4}{3} e^{3x}\).
Find the following integrals.
Integral: \(\int (4e^{3x} + 1) dx\)
Answer:
\(\int (4e^{3x} + 1) dx = 4 \int e^{3x} dx + \int 1 dx\)
\(= 4\left( \frac{e^{3x}}{3} \right) + x + C\)
\(= \frac{4}{3}e^{3x} + x + C\)
Find the following integrals.
Integral: \(\int x^2 (1 - \frac{1}{x^2}) dx\)
Answer:
\(\int x^2 (1 - \frac{1}{x^2}) dx\)
\(= \int (x^2 - 1) dx\)
\(= \int x^2 dx - \int 1 dx\)
\(= \frac{x^3}{3} - x + C\)
Find the following integrals.
Integral: \(\int (ax^2 + bx + c) dx\)
Answer:
\(\int (ax^2 + bx + c) dx\)
\(= a \int x^2 dx + b \int x dx + c \int 1 dx\)
\(= a\left( \frac{x^3}{3} \right) + b\left( \frac{x^2}{2} \right) + cx + C\)
\(= \frac{ax^3}{3} + \frac{bx^2}{2} + cx + C\)
Find the following integrals.
Integral: \(\int (2x^2 + e^x) dx\)
Answer:
\(\int (2x^2 + e^x) dx\)
\(= 2 \int x^2 dx + \int e^x dx\)
\(= 2\left( \frac{x^3}{3} \right) + e^x + C\)
\(= \frac{2x^3}{3} + e^x + C\)
Find the following integrals.
Integral: \(\int (\sqrt{x} - \frac{1}{\sqrt{x}})^2 dx\)
Answer:
\(\int (\sqrt{x} - \frac{1}{\sqrt{x}})^2 dx\)
Expanding using \((a-b)^2 = a^2 + b^2 - 2ab\):
\(= \int \left( x + \frac{1}{x} - 2 \right) dx\)
\(= \int x dx + \int \frac{1}{x} dx - 2 \int 1 dx\)
Now we know that, \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) and \(\int \frac{1}{x} dx = \log|x|\).
Therefore,
\(= \frac{x^2}{2} + \log|x| - 2x + C\)
Find the following integrals.
Integral: \(\int \frac{x^3 + 5x^2 - 4}{x^2} dx\)
Answer:
\(\int \frac{x^3 + 5x^2 - 4}{x^2} dx\)
Separating the terms we get,
\(= \int (x + 5 - 4x^{-2}) dx\)
Applying the formula, \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\)
\(= \int x dx + 5 \int 1 dx - 4 \int x^{-2} dx\)
\(= \frac{x^2}{2} + 5x - 4\left( \frac{x^{-1}}{-1} \right) + C\)
\(= \frac{x^2}{2} + 5x + \frac{4}{x} + C\)
Find the following integrals.
Integral: \(\int \frac{x^3 + 3x + 4}{\sqrt{x}} dx\)
Answer:
\(\int \frac{x^3 + 3x + 4}{\sqrt{x}} dx\)
Separating the terms we get,
\(= \int (x^{3 - \frac{1}{2}} + 3x^{1 - \frac{1}{2}} + 4x^{-\frac{1}{2}}) dx\)
\(= \int (x^{\frac{5}{2}} + 3x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}) dx\)
Applying the formula, \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\)
\(= \frac{x^{7/2}}{7/2} + 3\frac{x^{3/2}}{3/2} + 4\frac{x^{1/2}}{1/2} + C\)
\(= \frac{2}{7}x^{\frac{7}{2}} + 2x^{\frac{3}{2}} + 8x^{\frac{1}{2}} + C\)
\(= \frac{2}{7}x^{\frac{7}{2}} + 2x^{\frac{3}{2}} + 8\sqrt{x} + C\)
Find the following integrals.
Integral: \(\int \frac{x^3 - x^2 + x - 1}{x - 1} dx\)
Answer:
\(\int \frac{x^3 - x^2 + x - 1}{x - 1} dx\)
Now the numerator can be factorized as,
\(x^3 - x^2 + x - 1 = x^2(x - 1) + 1(x - 1)\)
\(x^3 - x^2 + x - 1 = (x^2 + 1)(x - 1)\)
Now putting this in given integral we get,
\(\int \frac{(x^2 + 1)(x - 1)}{x - 1} dx\)
\(= \int (x^2 + 1) dx\)
\(= \int x^2 dx + \int 1 dx\)
\(= \frac{x^3}{3} + x + C\)
Find the following integrals.
Integral: \(\int (1 - x)\sqrt{x} dx\)
Answer:
\(\int (1 - x)\sqrt{x} dx\)
\(= \int (\sqrt{x} - x\sqrt{x}) dx\)
\(= \int (x^{\frac{1}{2}} - x^{\frac{3}{2}}) dx\)
\(= \int x^{\frac{1}{2}} dx - \int x^{\frac{3}{2}} dx\)
\(= \frac{x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2} + C\)
\(= \frac{2}{3}x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}} + C\)
Find the following integrals.
Integral: \(\int \sqrt{x} (3x^2 + 2x + 3) dx\)
Answer:
\(\int \sqrt{x} (3x^2 + 2x + 3) dx\)
\(= \int (3x^{\frac{5}{2}} + 2x^{\frac{3}{2}} + 3x^{\frac{1}{2}}) dx\)
\(= 3 \int x^{\frac{5}{2}} dx + 2 \int x^{\frac{3}{2}} dx + 3 \int x^{\frac{1}{2}} dx\)
\(= 3\left( \frac{2}{7}x^{\frac{7}{2}} \right) + 2\left( \frac{2}{5}x^{\frac{5}{2}} \right) + 3\left( \frac{2}{3}x^{\frac{3}{2}} \right) + C\)
\(= \frac{6}{7}x^{\frac{7}{2}} + \frac{4}{5}x^{\frac{5}{2}} + 2x^{\frac{3}{2}} + C\)
Find the following integrals.
Integral: \(\int (2x - 3\cos x + e^x) dx\)
Answer:
\(\int (2x - 3\cos x + e^x) dx\)
\(= 2 \int x dx - 3 \int \cos x dx + \int e^x dx\)
\(= \frac{2x^2}{2} - 3(\sin x) + e^x + C\)
\(= x^2 - 3\sin x + e^x + C\)
Find the following integrals.
Integral: \(\int (2x^2 - 3\sin x + 5\sqrt{x}) dx\)
Answer:
\(\int (2x^2 - 3\sin x + 5\sqrt{x}) dx\)
\(= 2 \int x^2 dx - 3 \int \sin x dx + 5 \int x^{\frac{1}{2}} dx\)
\(= 2\left( \frac{x^3}{3} \right) - 3(-\cos x) + 5\left( \frac{2}{3}x^{\frac{3}{2}} \right) + C\)
\(= \frac{2x^3}{3} + 3\cos x + \frac{10}{3}x^{\frac{3}{2}} + C\)
Find the following integrals.
Integral: \(\int \sec x (\sec x + \tan x) dx\)
Answer:
Formulas Used: \(\int \sec^2 x dx = \tan x + C\) and \(\int \sec x \tan x dx = \sec x + C\)
\(\int \sec x (\sec x + \tan x) dx\)
Opening the brackets we get,
\(= \int (\sec^2 x + \sec x \tan x) dx\)
\(= \int \sec^2 x dx + \int \sec x \tan x dx\)
\(= \tan x + \sec x + C\)
Find the following integrals.
Integral: \(\int \frac{\sec^2 x}{\csc^2 x} dx\)
Answer:
\(\int \frac{\sec^2 x}{\csc^2 x} dx\)
\(= \int \frac{1/\cos^2 x}{1/\sin^2 x} dx\)
\(= \int \frac{\sin^2 x}{\cos^2 x} dx\)
\(= \int \tan^2 x dx\)
We know that \(\tan^2 x = \sec^2 x - 1\)
\(= \int (\sec^2 x - 1) dx\)
\(= \int \sec^2 x dx - \int 1 dx\)
\(= \tan x - x + C\)
Find the following integrals.
Integral: \(\int \frac{2 - 3\sin x}{\cos^2 x} dx\)
Answer:
\(\int \frac{2 - 3\sin x}{\cos^2 x} dx\)
\(= \int \left( \frac{2}{\cos^2 x} - \frac{3\sin x}{\cos^2 x} \right) dx\)
\(= \int (2\sec^2 x - 3\tan x \sec x) dx\)
\(= 2 \int \sec^2 x dx - 3 \int \tan x \sec x dx\)
\(= 2\tan x - 3\sec x + C\)
The anti-derivative of \((\sqrt{x} + 1/\sqrt{x})\) equals
A. \(\frac{1}{3} x^{1/3} + 2x^{1/2} + C\)
B. \(\frac{2}{3} x^{2/3} + \frac{1}{2} x^2 + C\)
C. \(\frac{2}{3} x^{3/2} + 2x^{1/2} + C\)
D. \(\frac{3}{2} x^{3/2} + \frac{1}{2} x^{1/2} + C\)
Answer:
\(\int (\sqrt{x} + \frac{1}{\sqrt{x}}) dx\)
\(= \int x^{\frac{1}{2}} dx + \int x^{-\frac{1}{2}} dx\)
\(= \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C\)
\(= \frac{2}{3}x^{\frac{3}{2}} + 2x^{\frac{1}{2}} + C\)
If \(\frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}\) such that \(f(2) = 0\). Then \(f(x)\) is
A. \(x^4 + \frac{1}{x^3} - \frac{129}{8}\)
B. \(x^3 + \frac{1}{x^4} + \frac{129}{8}\)
C. \(x^4 + \frac{1}{x^3} + \frac{129}{8}\)
D. \(x^3 + \frac{1}{x^4} - \frac{129}{8}\)
Solution:
It is given that \(\frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}\)
Integrating both sides:
\(f(x) = \int (4x^3 - \frac{3}{x^4}) dx\)
\(\Rightarrow f(x) = 4 \int x^3 dx - 3 \int x^{-4} dx\)
\(\Rightarrow f(x) = 4\left(\frac{x^4}{4}\right) - 3\left(\frac{x^{-3}}{-3}\right) + C\)
\(\Rightarrow f(x) = x^4 + \frac{1}{x^3} + C\)
Also, It is given that \(f(2) = 0\)
\(\Rightarrow f(2) = (2)^4 + \frac{1}{(2)^3} + C = 0\)
\(\Rightarrow 16 + \frac{1}{8} + C = 0\)
\(\Rightarrow C = - (16 + \frac{1}{8})\)
\(\Rightarrow C = \frac{-129}{8}\)
Therefore,
\(f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}\)
Title: NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.1
Labels: Class 12 Maths, Integrals, NCERT Solutions, CBSE
Permanent Link: class-12-maths-chapter-7-integrals-exercise-7-1-solution
Search Description: Step-by-step solutions for Class 12 Maths Exercise 7.1 Integrals. Detailed explanations, method of inspection, and anti-derivative calculations.
