NCERT Solutions for Class 12 Mathematics Chapter 7 Integrals

NCERT Solutions for Class 12 Mathematics Chapter 7 - Integrals

Welcome to the Chapter 7 - Integrals, Class 12 Mathematics NCERT Solutions page. Here, we provide detailed question answers for Chapter 7 - Integrals. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.

Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Integrals and excel in their exams. By going through these Integrals question answers, you can strengthen your foundation and improve your performance in Class 12 Mathematics. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.

Exercise 1

Q 1:

Integrals sin 2x

The anti derivative of sin 2x is a function of x whose derivative is sin 2x.

It is known that,

\begin{align} \frac {d}{dx} (cos 2x) = 2 sin2x \end{align}

⇒ \begin{align} sin 2x =-\frac {1}{2} \frac {d}{dx}(cos 2x) \end{align}

∴ \begin{align} sin 2x = \frac {d}{dx}\left(-\frac {1}{2}cos 2x\right) \end{align}

Therefore, the anti derivative of sin2x is

\begin{align} sin 2x \;is -\frac {1}{2}cos 2x \end{align}

Q 10:

\begin{align} \int \left(\sqrt{x} - \frac {1}{\sqrt{x}}\right)^2 .dx\end{align}

\begin{align} =\int \left(x + \frac {1}{x} - 2\right) .dx\end{align}

\begin{align} =\int x .dx + \int \frac {1}{x} . dx - 2\int 1. dx\end{align}

\begin{align} =\frac{x^2}{2} + \log\left(\left|x\right|\right) - 2x + C\end{align}

Q 11:

\begin{align} \int \frac{x^3 + 5x^2 - 4}{x^2} . dx\end{align}

\begin{align} =\int \left(x + 5 - 4x^{-2}\right) . dx\end{align}

\begin{align} =\int x .dx + 5 \int 1.dx- 4 \int x^{-2} .dx\end{align}

\begin{align} =\frac {x^2}{2} + 5x + 4 \left(\frac{x^{-1}}{-1}\right) + C\end{align}

\begin{align} =\frac {x^2}{2} + 5x + \frac{4}{x} + C\end{align}

Q 12:

\begin{align} \int \frac{x^3 + 3x + 4}{\sqrt{x}} . dx\end{align}

\begin{align} =\int \left(x^\frac{5}{2} + 3x^\frac{1}{2} + 4x^\frac{1}{2}\right) . dx\end{align}

\begin{align} =\frac{\left(x^{\displaystyle\frac72}\right)}{\displaystyle\frac72}+ \frac{3\left(x^{\displaystyle\frac32}\right)}{\displaystyle\frac32} + \frac{4\left(x^{\displaystyle\frac12}\right)}{\displaystyle\frac12} + C\end{align}

\begin{align} =\frac27\left(x^\frac72\right)+ 2\left(x^\frac32\right) + 8\left(x^\frac12\right) + C\end{align}

\begin{align} =\frac27\left(x^\frac72\right)+ 2\left(x^\frac32\right) + 8\left(\sqrt x\right) + C\end{align}

Q 13:

\begin{align} \int \frac{x^3 - x^2 + x - 1}{x-1} . dx\end{align}

On dividing, we obtain

\begin{align} =\int \left({x^2 + 1}\right) . dx \end{align}

\begin{align} =\int {x^2} . dx + \int 1 .dx \end{align}

\begin{align} =\frac {x^3}{3} + x + C \end{align}

Q 14:

\begin{align} \int\left(1-x\right).\sqrt {x}.dx\end{align}

\begin{align} =\int \left(\sqrt {x} - x^\frac32\right).dx\end{align}

\begin{align} =\frac{x^{\displaystyle\frac32}}{\displaystyle\frac32} - \frac{x^{\displaystyle\frac52}}{\displaystyle\frac52}+C \end{align}

\begin{align} =\frac23\;x^\frac32\; - \frac25\;x^\frac52\;+C \end{align}

Q 15:

\begin{align} \int\sqrt {x}.\left(3x^2+2x + 3\right).dx\end{align}

\begin{align} =\int\left(3x^\frac52+2x^\frac32 + 3x^\frac12\right).dx\end{align}

\begin{align} =3\int x^\frac52 .dx+2 \int x^\frac32 .dx + 3\int x^\frac12.dx\end{align}

\begin{align} =3\left(\frac{x^{\displaystyle\frac72}}{\displaystyle\frac72}\right) + 2\left(\frac{x^{\displaystyle\frac52}}{\displaystyle\frac52}\right)+3\left(\frac{x^{\displaystyle\frac32}}{\displaystyle\frac32}\right)+C \end{align}

\begin{align} =\frac67\;x^\frac72\; + \frac45\;x^\frac52\;+ 2x^\frac32\; +C \end{align}

Q 16:

\begin{align} \int\left(2x - 3Cosx + e^x\right).dx\end{align}

\begin{align} =2\int x .dx - 3\int Cosx .dx + \int e^x.dx\end{align}

\begin{align} =\frac{2x^2}{2} - 3(Sinx) + e^x + C\end{align}

\begin{align} =x^2 - 3Sinx + e^x + C\end{align}

Q 17:

\begin{align} \int\left(2x^2-3Sinx +5\sqrt {x}\right).dx\end{align}

\begin{align} =2\int x^2.dx-3\int Sinx.dx +5\int x^\frac12.dx \end{align}

\begin{align} =\frac{2x^3}{3} - 3(- Cos x) +5\left(\frac{x^{\displaystyle\frac32}}{\displaystyle\frac32}\right) + C \end{align}

\begin{align} =\frac{2}{3}.x^3 + 3Cos x + \frac{10}{3}.x^\frac{3}{2} + C\end{align}

Q 18:

\begin{align} \int sec x . \left(sec x + tan x\right) .dx \end{align}

\begin{align} =\int \left(sec^2 x + sec x . tan x\right) .dx \end{align}

\begin{align} =\int sec^2 x . dx + \int sec x . tan x .dx \end{align}

\begin{align} = tan x + sec x + C \end{align}

Q 19:

\begin{align} \int \frac {sec^2 x}{Coses^2 x} . dx\end{align}

\begin{align} =\int \left(\frac{\frac {1}{Cos^2 x}}{\frac{1}{sin^2 x}}\right) . dx\end{align}

\begin{align} =\int \left(\frac{Sin^2x}{Cos^2x}\right) . dx\end{align}

\begin{align} =\int tan^2 x . dx\end{align}

\begin{align} =\int \left(sec^2x - 1\right) . dx\end{align}

\begin{align} =\int sec^2x . dx - \int 1. dx\end{align}

\begin{align} = tanx - x + C\end{align}

Q 2:

Integrals cos3x

The anti derivative of cos 3x is a function of x whose derivative is cos 3x.

It is known that,

\begin{align} \frac {d}{dx} (sin 3x) = 3 cos3x \end{align}

⇒ \begin{align} cos 3x =\frac {1}{3} \frac {d}{dx}(sin 3x) \end{align}

∴ \begin{align} cos 3x = \frac {d}{dx}\left(\frac {1}{3}sin 3x\right) \end{align}

Therefore, the anti derivative of cos3x is

\begin{align} sin 2x \;is \frac {1}{3}sin 3x \end{align}

Q 20:

\begin{align} \int \left(\frac {2-3sin x}{cos^2 x}\right) . dx\end{align}

\begin{align} =\int \left(\frac {2}{cos^2 x} - \frac{3sinx}{cos^2x}\right) . dx\end{align}

\begin{align} =2\int sec^2 x .dx - 3 \int tan x . sec x. dx\end{align}

\begin{align} =2 tan x - 3 sec x + C\end{align}

Q 21:

The anti derivative of \begin{align} \left(\sqrt x + \frac {1}{\sqrt x}\right)\end{align} equals to \begin{align} (A) \frac{1}{3}.x^\frac{1}{3} + 2.x^\frac{1}{2} +C \;\;\;\; (B) \frac{2}{3}.x^\frac{2}{3} + \frac{1}{2}.x^2 +C \end{align}
\begin{align} (C) \frac{2}{3}.x^\frac{3}{2} +2 x^\frac{1}{2} +C \;\;\;\;(D) \frac{3}{2}.x^\frac{3}{2} +\frac{1}{2}. x^\frac{1}{2} +C \end{align}

\begin{align} \int\left(\sqrt x + \frac {1}{\sqrt x}\right).dx \end{align}

\begin{align}= \int x^\frac{1}{2}.dx + \int x^\frac{-1}{2}.dx \end{align}

\begin{align}= \left(\frac {x^\frac{3}{2}}{\frac{3}{2}}\right) + \left(\frac {x^\frac{1}{2}}{\frac{1}{2}}\right) + C\end{align}

\begin{align}= \frac {2}{3} . x^{\frac{3}{2}}+ 2x^\frac{1}{2} + C\end{align}

Hence, the Correct Answer is C.

Q 22:

If \begin{align} \frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}\end{align} such that f(2) = 0 , then f(x) is
\begin{align} (A) x^4 + \frac {1}{x^3} - \frac{129}{8} \;\;\;\;(B) x^3 + \frac{1}{x^4} + \frac{129}{8}\end{align}
\begin{align} (c) x^3 + \frac {1}{x^4} + \frac{129}{8} \;\;\;\;(D) x^3 + \frac{1}{x^4} - \frac{129}{8}\end{align}

It is given that,

\begin{align} \frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}\end{align}

∴ Anti derivative of

\begin{align} 4x^3 - \frac{3}{x^4} = f(x)\end{align}

∴ \begin{align} f(x)= \int \left(4x^3 - \frac{3}{x^4}\right).dx\end{align}

\begin{align} f(x)= 4\int x^3.dx - 3\int {x^{-4}}.dx\end{align}

\begin{align} f(x)= 4\left(\frac {x^4}{4}\right) - 3\left(\frac {x^{-3}}{-3}\right) + C\end{align}

∴ \begin{align} f(x)= x^4 + \frac{1}{x^3} + C\end{align}

Also, f(2) = 0

∴ \begin{align} f(2) =\left(2\right)^4 + \frac{1}{\left(2\right)^3} + C = 0 \end{align}

=> \begin{align} 16 + \frac{1}{8} + C = 0 \end{align}

=> \begin{align} C = -\left(16 + \frac{1}{8}\right) \end{align}

=> \begin{align} C = \frac{-129}{8} \end{align}

∴ \begin{align} f(x)= x^4 + \frac{1}{x^3} -\frac{129}{8} \end{align}

Hence, the correct answer is A.

Q 3:

Integrals e^2x

The anti derivative of e^2x is a function of x whose derivative is e^2x.

It is known that,

\begin{align} \frac {d}{dx} (e^{2x}) = 2e^{2x} \end{align}

⇒ \begin{align} e^{2x} =\frac {1}{2} \frac {d}{dx}(e^{2x}) \end{align}

∴ \begin{align} e^{2x} = \frac {d}{dx}\left(\frac {1}{2}e^{2x}\right) \end{align}

Therefore, the anti derivative of e^2x

\begin{align} e^{2x} \;is \frac {1}{2}e^{2x} \end{align}

Q 4:

Integrals (ax + b)²

The anti derivative of (ax + b)² is a function of x whose derivative is (ax + b)².

It is known that,

\begin{align} \frac {d}{dx} ((ax+b)^3) = 3a(ax+b)^2 \end{align}

⇒ \begin{align} (ax + b)^2 =\frac {1}{3a} \frac {d}{dx}(ax+b)^3 \end{align}

∴ \begin{align} (ax + b)^2 = \frac {d}{dx}\left(\frac {1}{3a}(ax + b)^3\right) \end{align}

Therefore, the anti derivative of (ax +b)²

\begin{align} (ax + B)^2 \;is \frac {1}{3a}(ax +b)^3 \end{align}

Q 5:

sin 2x – 4e^3x

The anti derivative of sin 2x – 4e^3x is the function of x whose derivative is sin 2x – 4e^3x.

It is known that,

\begin{align} \frac {d}{dx} \left(-\frac{1}{2}cos 2x – \frac {4}{3} e^{3x}\right) = sin2x – 4e^{3x} \end{align}

Therefore, the anti derivative of (sin 2x – 4e^3x) is \begin{align} \left(-\frac{1}{2}cos 2x – \frac {4}{3} e^{3x}\right) \end{align}

Q 6:

\begin{align} \int \left(4e^{3x} + 1\right).dx \end{align}

\begin{align} =4\int e^{3x}.dx + \int 1.dx \end{align}

\begin{align} =4\left(\frac {e^{3x}}{3}\right) + x + C \end{align}

\begin{align} =\frac {4}{3} e^{3x} + x + C \end{align}

Q 7:

\begin{align} \int x^2\left(1 - \frac{1}{x^2}\right).dx \end{align}

\begin{align} =\int \left(x^2 - 1\right).dx \end{align}

\begin{align} =\int x^2.dx - \int1.dx \end{align}

\begin{align} =\frac{x^3}{3} - x + C \end{align}

Q 8:

\begin{align} \int \left({a}{x^2} + bx + c\right) .dx\end{align}

\begin{align} =a\int {x^2}.dx + b\int x.dx + c\int 1.dx \end{align}

\begin{align} =a\left(\frac {x^3}{3}\right) + b\left(\frac {x^2}{2}\right) + cx + C\end{align}

Q 9:

\begin{align} \int \left({2}{x^2} + e^x\right) .dx\end{align}

\begin{align} =2\int {x^2}.dx + \int e^x.dx \end{align}

\begin{align} =2\left(\frac {x^3}{3}\right) +e^x + C\end{align}

\begin{align} =\frac {2}{3}.x^3 +e^x + C\end{align}

Frequently Asked Questions about Integrals - Class 12 Mathematics

- 1. How many questions are covered in Integrals solutions?
- All questions from Integrals are covered with detailed step-by-step solutions including exercise questions, additional questions, and examples.
- 2. Are the solutions for Integrals helpful for exam preparation?
- Yes, the solutions provide comprehensive explanations that help students understand concepts clearly and prepare effectively for both board and competitive exams.
- 3. Can I find solutions to all exercises in Integrals?
- Yes, we provide solutions to all exercises, examples, and additional questions from Integrals with detailed explanations.
- 4. How do these solutions help in understanding Integrals concepts?
- Our solutions break down complex problems into simple steps, provide clear explanations, and include relevant examples to help students grasp the concepts easily.
- 5. Are there any tips for studying Integrals effectively?
- Yes, practice regularly, understand the concepts before memorizing, solve additional problems, and refer to our step-by-step solutions for better understanding.

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The Integrals is an important chapter of 12 Mathematics. This chapter’s important topics like Integrals are often featured in board exams. Practicing the question answers from this chapter will help you rank high in your board exams.

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