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NCERT Solutions for Exercise 7.6 Class 12 Maths Chapter 7 - Integrals

NCERT Solutions for Exercise 7.6 Class 12 Maths Chapter 7 - Integrals

Edited By Ramraj Saini | Updated on Dec 03, 2023 10:01 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.6

NCERT Solutions for Exercise 7.6 Class 12 Maths Chapter 7 Integrals are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. In NCERT solutions for Class 12 Maths chapter 7 exercise 7.6 functions like logarithmic, inverse trigonometric functions, exponential functions etc. are discussed in step by step manner. Solutions to exercise 7.6 Class 12 Maths are prepared in a holistic manner by subject matter experts. Hence NCERT solutions for Class 12 Maths chapter 7 exercise 7.6 provided below can be a good source. There is no need to refer to any other book for these topics as the content provided in the NCERT book is more than sufficient for the examination. Practicing from this exercise can help students score well in exams like JEE Main.

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Access NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.6

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Integrals Class 12 Chapter 7 Exercise: 7.6

Question:1 Integrate the functions x \sin x

Answer:

Given function is
f(x)=x \sin x
We will use integrate by parts method
\int x\sin x = x.\int \sin xdx - \int(\frac{d(x)}{dx}.\int sin x dx)dx\\ \\ \int x\sin x = x.(-\cos x)- \int (1.(-\cos x))dx\\ \\ \int x\sin x= -x\cos x+\sin x + C
Therefore, the answer is -x\cos x+\sin x + C

Question:2 Integrate the functions x \sin 3x

Answer:

Given function is
f(x)=x \sin 3x
We will use integration by parts method
\int x\sin 3x = x.\int \sin 3xdx - \int(\frac{d(x)}{dx}.\int sin 3x dx)dx\\ \\ \int x\sin 3x = x.(\frac{-\cos 3x}{3})- \int (1.(\frac{-\cos 3x}{3}))dx\\ \\ \int x\sin 3x= -\frac{x\cos 3x}{3}+\frac{\sin 3x}{9} + C

Therefore, the answer is -\frac{x\cos 3x}{3}+\frac{\sin 3x}{9} + C

Question:3 Integrate the functions x ^ 2 e ^x

Answer:

Given function is
f(x)=x^2e^x
We will use integration by parts method
\int x^2e^x= x^2.\int e^xdx - \int(\frac{d(x^2)}{dx}.\int e^x dx)dx\\ \\ \int x^2e^x = x^2.e^x- \int (2x.e^x)dx\\
Again use integration by parts in \int (2x.e^x)dx\\
\int (2x.e^x)dx = 2x.\int e^x dx - \int (\frac{d(2x)}{dx}.\int e^xdx)dx\\ \int 2x.e^x dx = 2xe^x- \int 2.e^xdx\\ \int 2x.e^x dx = 2xe^x- 2e^x
Put this value in our equation
we will get,
\int x^2.e^x dx =x^2e^x -2xe^x+ 2e^x + C\\ \int x^2.e^x dx = e^x(x^2-2x+2)+ C

Therefore, answer is e^x(x^2-2x+2)+ C

Question:4 Integrate the functions x \log x

Answer:

Given function is
f(x)=x.\log x
We will use integration by parts method
\int x.\log xdx= \log x.\int xdx - \int(\frac{d(\log x)}{dx}.\int x dx)dx\\ \\ \int x\log xdx = \log x.\frac{x^2}{2}- \int (\frac{1}{x}.\frac{x^2}{2})dx\\ \int x\log xdx = \log x.\frac{x^2}{2}- \int \frac{x}{2}dx\\ \int x\log xdx = \log x.\frac{x^2}{2}- \frac{x^2}{4}+ C

Therefore, the answer is \frac{x^2}{2}\log x- \frac{x^2}{4}+ C

Question:5 Integrate the functions x \log 2x

Answer:

Given function is
f(x)=x.\log 2 x
We will use integration by parts method
\int x.\log 2xdx= \log 2x.\int xdx - \int(\frac{d(\log 2x)}{dx}.\int x dx)dx\\ \\ \int x\log 2xdx = \log 2x.\frac{x^2}{2}- \int (\frac{2}{2x}.\frac{x^2}{2})dx\\ \int x\log 2xdx = \log 2x.\frac{x^2}{2}- \int \frac{x}{2}dx\\ \int x\log 2xdx = \log 2x.\frac{x^2}{2}- \frac{x^2}{4}+ C

Therefore, the answer is \log 2x.\frac{x^2}{2}- \frac{x^2}{4}+ C

Question:6 Integrate the functions x^ 2 \log x

Answer:

Given function is
f(x)=x^2.\log x
We will use integration by parts method
\int x^2.\log xdx= \log x.\int x^2dx - \int(\frac{d(\log x)}{dx}.\int x^2 dx)dx\\ \\ \int x^2\log xdx = \log x.\frac{x^3}{3}- \int (\frac{1}{x}.\frac{x^3}{3})dx\\ \int x^2\log xdx = \log x.\frac{x^3}{3}- \int \frac{x^2}{3}dx\\ \int x^2\log xdx = \log x.\frac{x^3}{3}- \frac{x^3}{9}+ C

Therefore, the answer is \log x.\frac{x^3}{3}- \frac{x^3}{9}+ C

Question:7 Integrate the functions x \sin ^{ -1} x

Answer:

Given function is
f(x)=x.\sin^{-1} x
We will use integration by parts method
\int x.\sin^{-1} xdx= \sin^{-1} x.\int xdx - \int(\frac{d(\sin^{-1} x)}{dx}.\int x dx)dx\\ \\ \int x\sin^{-1} xdx = \sin^{-1} x.\frac{x^2}{2}- \int (\frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\
Now, we need to integrate \int (\frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\
\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\left ( \int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{1}{2}\left ( \frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x-\sin^{-1}x \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{x\sqrt{1-x^2}}{4} -\frac{\sin^{-1}x}{4}+C
Put this value in our equation

Therefore, the answer is \int x\sin^{-1} xdx =\frac{\sin^{-1}x}{4}(2x^2-1)-\frac{x\sqrt{1-x^2}}{4}

Question:8 Integrate the functions x \tan ^{-1} x

Answer:

Given function is
f(x)=x.\tan^{-1} x
We will use integration by parts method
\int x.\tan^{-1} xdx= \tan^{-1} x.\int xdx - \int(\frac{d(\tan^{-1} x)}{dx}.\int x dx)dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}- \int (\frac{1}{1+x^2}.\frac{x^2}{2})dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\int \left ( \frac{x^2+1}{1+x^2}-\frac{1}{1+x^2} \right )dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\int \left ( 1-\frac{1}{1+x^2} \right )dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\left ( x- \tan^{-1}x \right )+C\\ \\ \int x\tan^{-1}xdx = \frac{\tan^{-1}x}{2}(2x^2+1)-\frac{x}{2}+C

Put this value in our equation
\int x\sin^{-1} xdx = \sin^{-1} x.\frac{x^2}{2}- \frac{x}{4\sqrt{1-x^2}}-\frac{\sin^{-1}x}{4}+C\\ \int x\sin^{-1} xdx =\frac{\sin^{-1}x}{4}(2x^2-1)-\frac{x}{4\sqrt{1-x^2}}

Therefore, the answer is \frac{\tan^{-1}x}{2}(2x^2+1)-\frac{x}{2}+C

Question:9 Integrate the functions x\cos ^{ -1} x

Answer:

Given function is
f(x)=x.\cos^{-1} x
We will use integration by parts method
\int x.\cos^{-1} xdx= \cos^{-1} x.\int xdx - \int(\frac{d(\cos^{-1} x)}{dx}.\int x dx)dx\\ \\ \int x\cos^{-1} xdx = \cos^{-1} x.\frac{x^2}{2}- \int (\frac{-1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\
Now, we need to integrate \int (\frac{-1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\
\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\left ( \int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{1}{2}\left ( \frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}\cos^{-1}x+\cos^{-1}x \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{x\sqrt{1-x^2}}{4} -\frac{\cos^{-1}x}{4}+\frac{\cos^{-1}x}{2}+C
Put this value in our equation
\int x\cos^{-1} xdx = \cos^{-1} x.\frac{x^2}{2}-\left ( \frac{x\sqrt{1-x^2}}{4} -\frac{\cos^{-1}x}{4}+\frac{\cos^{-1}x}{2}+C \right )\\ \\ \int x\cos^{-1} xdx =\frac{\cos^{-1}x}{4}(2x^2-1)- \frac{x\sqrt{1-x^2}}{4}

Therefore, the answer is \frac{\cos^{-1}x}{4}(2x^2-1)- \frac{x\sqrt{1-x^2}}{4}

Question:10 Integrate the functions ( \sin ^{-1}x ) ^ 2

Answer:

Given function is
f(x)=( \sin ^{-1}x ) ^ 2
we will use integration by parts method
\int (\sin^{-1}x)^2= (\sin^{-1}x)^2.\int 1dx-\int \left ( \frac{d( (\sin^{-1}x)^2)}{dx} .\int 1dx\right )dx\\ \\ \int (\sin^{-1}x)^2 = (\sin^{-1}x)^2.x-\int \left ( \sin^{-1}.\frac{2x}{\sqrt{1-x^2}} \right )dx\\ \int (\sin^{-1}x)^2 = (\sin^{-1}x)^2.x + \left [ \sin^{-1}x.\int \frac{-2x}{\sqrt{1-x^2}}dx-\int \left ( \frac{d(\sin^{-1}x)}{dx}. \int \frac{-2x}{\sqrt{1-x^2}}dx\right ) \right ]\\ \\ . \ \ \ \ \ = (\sin^{-1}x)^2.x + \left [ \sin^{-1}x.2\sqrt{1-x^2}- \int \frac{1}{\sqrt{1-x^2}}.2\sqrt{1-x^2}dx \right ]\\ \\ . \ \ \ \ \ = (\sin^{-1}x)^2.x + 2\sin^{-1}x\sqrt{1-x^2}-2x+C
Therefore, answer is (\sin^{-1}x)^2.x + 2\sin^{-1}x\sqrt{1-x^2}-2x+C

Question:11 Integrate the functions \frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}

Answer:

Consider \int \frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}dx =I

So, we have then: I = \frac{-1}{2}\int \frac{-2x}{\sqrt{1-x^2}}. \cos^{-1}x dx

After taking \cos ^{-1}x as a first function and \left ( \frac{-2x}{\sqrt{1-x^2}} \right ) as second function and integrating by parts, we get

I =-\frac{1}{2}\left [ \cos^{-1}x\int\frac{-2x}{\sqrt{1-x^2}}dx - \int\left \{ \left ( \frac{d}{dx}\cos^{-1}x \right )\int \frac{-2x}{\sqrt{1-x^2}}dx \right \}dx \right ] =-\frac{1}{2}\left [ \cos^{-1}x.2{\sqrt{1-x^2}} + \int \frac{-1}{\sqrt{1-x^2}}.2\sqrt{1-x^2}dx \right ]

=\frac{-1}{2}\left [ 2\sqrt{1-x^2}\cos^{-1}x-\int2dx \right ]

=\frac{-1}{2}\left [ 2\sqrt{1-x^2}\cos^{-1}x-2x \right ]+C

Or, -\left \sqrt{1-x^2}\cos^{-1}x +x\right +C

Question:12 Integrate the functions x \sec ^2 x

Answer:

Consider x \sec ^2 x

So, we have then: I =\int x\sec^2 x dx

After taking x as a first function and \sec^2x as second function and integrating by parts, we get

I =x\int \sec^2 x dx -\int \left \{ \left ( \frac{d}{dx}x \right )\int \sec^2 x dx \right \}dx

= x\tan x -\int1.\tan x dx

= x\tan x +\log|\cos x | +C

Question:13 Integrate the functions \tan ^{-1} x

Answer:

Consider \tan ^{-1} x

So, we have then: I =\int 1.\tan^{-1}x dx

After taking \tan^{-1}x as a first function and 1 as second function and integrating by parts, we get

I = \tan^{-1}x \int 1dx -\int \left \{ \left ( \frac{d}{dx}\tan^{-1}x \right )\int1.dx \right \}dx

= \tan^{-1}x.x -\int \frac{1}{1+x^2}.xdx

= x\tan^{-1}x -\frac{1}{2}\int \frac{2x}{1+x^2}dx

= x\tan^{-1}x -\frac{1}{2}\log|1+x^2|+C

= x\tan^{-1}x -\frac{1}{2}\log(1+x^2)+C

Question:14 Integrate the functions x ( \log x )^ 2

Answer:

Consider x ( \log x )^ 2

So, we have then: I = \int x(\log x)^2 dx

After taking (\log x )^2 as a first function and x as second function and integrating by parts, we get

I = (\log x )^2 \int xdx -\int \left \{ \left ( \frac{d}{dx} (\log x)^2 \right )\int x.dx \right \}dx

= (\log x)^2 .\frac{x^2}{2} - \int \frac{2\log x }{x}.\frac{x^2}{2} dx

= (\log x)^2 .\frac{x^2}{2} - \int x\log x dx

= (\log x)^2 .\frac{x^2}{2} - \left ( \frac{x^2 \log x }{2} -\frac{x^2}{4} \right )+C

Question:15 Integrate the functions ( x^2 + 1 ) \log x

Answer:

Consider ( x^2 + 1 ) \log x

So, we have then: I = \int (x^2+1) \log x dx = \int x^2 \log x dx +\int \log x dx

Let us take I = I_{1} +I_{2} ....................(1)

Where, I_{1} = \int x^2\log x dx and I_{2} = \int \log x dx

So, I_{1} = \int x^2\log x dx

After taking \log x as a first function and x^2 as second function and integrating by parts, we get

I = \log x \int x^2dx -\int \left \{ \left ( \frac{d}{dx} \log x \right )\int x^2.dx \right \}dx

= \log x .\frac{x^3}{3} - \int \frac{1}{x}.\frac{x^3}{3} dx

= \log x .\frac{x^3}{3} - \frac{x^3}{9} +C_{1} ....................(2)

I_{2} = \int \log x dx

After taking \log x as a first function and 1 as second function and integrating by parts, we get

I_{2} = \log x \int 1.dx - \int \left \{ \left ( \frac{d}{dx}\log x \right ) \int 1.dx \right \}dx

= \log x .x -\int \frac{1}{x}. xdx

= x\log x -\int 1 dx

= x\log x -x +C_{2} ................(3)

Now, using the two equations (2) and (3) in (1) we get,

I = \frac{x^3}{3}\log x -\frac{x^3}{9} +C_{1} +x\log x - x +C_{2}

= \frac{x^3}{3}\log x -\frac{x^3}{9} +x\log x - x +(C_{1}+C_{2})

=\left ( \frac{x^3}{3}+x \right ) \log x -\frac{x^3}{9} -x+C

Question:16 Integrate the functions e ^ x ( \sin x + \cos x )

Answer:

Let suppose
I = e ^ x ( \sin x + \cos x )
f(x) = \sin x \Rightarrow f'(x) = \cos x
we know that,
I =\int e^x[f(x)+f'(x)]dx = e^x[f(x)]+C
Thus, the solution of the given integral is given by

\therefore I = e^x\sin x +C

Question:17 Integrate the functions \frac{x e ^x }{( 1+ x )^2}

Answer:

\frac{x e ^x }{( 1+ x )^2}
Let suppose
I = \int \frac{e^x(x)}{(1+x)^2}dx
by rearranging the equation, we get
\Rightarrow \int e^x[\frac{1}{1+x}-\frac{1}{(1+x)^2}]dx
let
f(x)=\frac{1}{1+x} \Rightarrow f'(x)= -\frac{1}{(1+x)^2}
It is known that \int e^x[f(x)+f'(x)]=e^x[f(x)]+C
therefore the solution of the given integral is

I = \frac{e^x}{1+x}+C

Question:18 Integrate the functions e ^x \left ( \frac{1+ \sin x }{1+ \cos x } \right )

Answer:

Let
I =e ^x \left ( \frac{1+ \sin x }{1+ \cos x } \right )
substitute 1 =\sin ^2\frac{x}{2}+\cos^2\frac{x}{2} and \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}

\\\Rightarrow e^x(\frac{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}})\\ =e^x(\frac{1}{2}\sec^2\frac{x}{2}+\tan\frac{x}{2})\\
let
f(x) =\tan\frac{x}{2} \Rightarrow f'(x)=\frac{1}{2}\sec^2\frac{x}{2}
It is known that \int e^x[f(x)+f'(x)]=e^x[f(x)]+C
Therefore the solution of the given integral is

I = e^x\tan\frac{x}{2} +C

Question:19 Integrate the functions e ^ x \left ( \frac{1 }{x} - \frac{1}{x^2}\right )

Answer:

e ^ x \left ( \frac{1 }{x} - \frac{1}{x^2}\right )
It is known that
\int e^x[f(x)+f'(x)]=e^x[f(x)]+C

let
f(x)=\frac{1}{x}\Rightarrow f'(x)=-\frac{1}{x^2}
Therefore the required solution of the given above integral is
I = e^x.\frac{1}{x}+C

Question:20 Integrate the functions \frac{( x-3)e ^x }{( x-1)^3}

Answer:

\frac{( x-3)e ^x }{( x-1)^3}
It is known that \int e^x[f(x)+f'(x)]=e^x[f(x)]+C

So, By adjusting the given equation, we get
\int\frac{( x-3)e ^x }{( x-1)^3} =\int e^x(\frac{x-1-2}{(x-1)^3}) =\int e^x({\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3})}dx

to let
f(x)=\frac{1}{(x-1)^2}\Rightarrow f'(x)=-\frac{2}{(x-1)^3}
Therefore the required solution of the given I=\frac{e^x}{(x-1)^2}+C integral is

Question:21 Integrate the functions e ^{ 2x } \sin x

Answer:

Let
I =e ^{ 2x } \sin x
By using integrating by parts, we get

\\=\sin x\int e ^{ 2x }dx-\int(\frac{d}{dx}\sin x.\int e^{2x}dx)\ dx\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{2}\int e^{2x}.\cos x\ dx\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{2}[\cos x\int e^{2x}dx-\int (\frac{d}{dx}\cos x.\int e^{2x}dx)\ dx]\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{2}[\cos x.\frac{e^x}{2}+\frac{1}{2}\int e^{2x}\sin x dx]\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{4}\cos x.e^{2x}-\frac{1}{4}I\\ \Rightarrow \frac{5}{4}I =\frac{\sin x.e^{2x}}{2}-\frac{1}{4}\cos x.e^{2x}\\ I = \frac{e^{2x}}{5}[2\sin x-\cos x]+C

Question:22 Integrate the functions \sin ^ { -1} \left ( \frac{2x}{1+x^2 } \right )

Answer:

\sin ^ { -1} \left ( \frac{2x}{1+x^2 } \right )

\int \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx
let x = \tan\theta\Rightarrow dx =\sec^2\theta d\theta

\\=\int\sin^{-1}(\frac{2\tan\theta}{1+\tan\theta})\sec^2\theta d\theta\\ =\int\sin^{-1}(\sin 2\theta)\sec^2\theta d\theta\\ =\int2\theta \sec^2\theta d\theta\\
Taking \theta as a first function and \sec^2\theta as a second function, by using by parts method

\\=2[\theta\int \sec^2\theta d\theta-\int(\frac{d}{d\theta}\theta.\int \sec^2\theta\ d\theta)d\theta]\\ =2[\theta\tan\theta-\int \tan\theta\ d\theta]+C\\ =2[\theta\tan\theta+\log\left | \cos\theta \right |]+C\\ =2x\tan^{-1}x+2\log (1+x^2)^{-1/2}\\ =2x\tan^{-1}x-\log(1+x^2)+C

Question:23 Choose the correct answer

\int x ^ 2 e ^{x ^3 } dx \: \: equals

A ) \frac{1}{3} e ^{x^3} + C \\\\ B) \frac{1}{3} e ^{x^2} + C \\\\ C ) \frac{1}{2} e ^{x^3} + C \\\\ D ) \frac{1}{2} e ^{x^2} + C

Answer:

the integration can be done ass follows

let x^3 =t\Rightarrow 3x^2dx=dt
\Rightarrow I =\frac{1}{3}\int e^tdt =\frac{1}{3}e^t+C=\frac{1}{3}e^x^3+C

Question:24 Choose the correct answer

\int e ^ x \sec ( 1+ \tan x ) dx \: \: \: equals

A ) e ^ x \cos x + C \\\\ B) e ^ x \sec x + C \\\\ C ) e ^ x \sin x + C\\\\D ) e ^ x \tan x + C

Answer:

we know that,
I =\int e^x[f(x)+f'(x)]dx = e^x[f(x)]+C
from above integral
let
f(x)=\sec x\Rightarrow f'(x)= \sec x.\tan x
thus, the solution of the above integral is
I=e^x\sec x+C

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There are broadly 2 types of integration, i.e Definite and indefinite Integrals. 

2. What are indefinite integrals ?

Indefinite integrals are defined without upper and lower limits i.e its range is not defined. 

3. What are the applications of Integrals ?

Integrals can be used in finding the quantities of area, volume, displacement etc. 

4. Are questions given here very difficult ?

This exercise mainly deals with advanced level of problems which are important for competitive examinations only. In the board examination, 5 markers can be asked from this exercise. 

5. Mention some of the topics in Exercise 7.6 Class 12 Maths.

Integrals of logarithmic, exponential and inverse trigonometric functions are discussed in this exercise. 

6. Mention the total number of questions in this Exercise 7.6 Class 12 Maths.

There are 24 questions in this Exercise 7.6 Class 12 Maths.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer
Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Read Complete Answer
Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read More

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
Read More

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read More

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read More

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
Read More

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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