NCERT Solutions for Exercise 7.9 Class 12 Maths Chapter 7

NCERT Solutions for Exercise 7.9 Class 12 Maths Chapter 7 - Integrals

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

NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.9

NCERT Solutions for Exercise 7.9 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 7.9 is one of the most important exercises from the exam perspective as it deals with the evaluation of definite integrals in a defined range. Such NCERT book questions are more often seen in the Board as well as competitive examination. Solutions to exercise 7.9 Class 12 Maths which are provided here are prepared in detail by the experienced subject matter experts. Considering the importance of NCERT solutions for Class 12 Maths chapter 7 exercise 7.9, it is highly recommended to students to practice at least a few questions before the examination.

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Integrals Class 12 Chapter 7 Exercise 7.9

Question:1 Evaluate the definite integrals in Exercises 1 to 20.

$\int_{-1}^{1} (x+1)dx$

Answer:

Given integral: $I = \int_{-1}^{1} (x+1)dx$

Consider the integral $\int (x+1)dx$

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

So, we have the function of $x$ , $f(x) = \frac{x^2}{2}+x$

Now, by Second fundamental theorem of calculus, we have

$I = f(1)-f(-1)$

$= \left ( \frac{1}{2}+1\right ) - \left (\frac{1}{2}-1 \right ) = \frac{1}{2}+1-\frac{1}{2}+1 = 2$

Question:2 Evaluate the definite integrals in Exercises 1 to 20.

$\int_2^3\frac{1}{x}dx$

Answer:

Given integral: $I = \int_2^3\frac{1}{x}dx$

Consider the integral $\int_2^3\frac{1}{x}dx$

$\int \frac{1}{x}dx = \log|x|$

So, we have the function of $x$ , $f(x) = \log|x|$

Now, by Second fundamental theorem of calculus, we have

$I = f(3)-f(2)$

$=\log|3|-\log|2| = \log \frac{3}{2}$

Question:3 Evaluate the definite integrals in Exercises 1 to 20.

$\int_1^2(4x^3-5x^2 + 6x +9)dx$

Answer:

Given integral: $I = \int_1^2(4x^3-5x^2 + 6x +9)dx$

Consider the integral $I = \int (4x^3-5x^2 + 6x +9)dx$

$\int (4x^3-5x^2 + 6x +9)dx = 4\frac{x^4}{4} -5\frac{x^3}{3}+6\frac{x^2}{2}+9x$

$= x^4 -\frac{5x^3}{3}+3x^2+9x$

So, we have the function of $x$ , $f(x) = x^4 -\frac{5x^3}{3}+3x^2+9x$

Now, by Second fundamental theorem of calculus, we have

$I = f(2)-f(1)$

$=\left \{ 2^4-\frac{5(2)^3}{3}+3(2)^2+9(2)\right \} - \left \{ 1^4-\frac{5(1)^3}{3}+3(1)^2+9(1) \right \}$

$=\left \{ 16-\frac{40}{3}+12+18\right \} - \left \{ 1-\frac{5}{3}+3+9 \right \}$

$=\left \{ 46-\frac{40}{3}\right \} - \left \{ 13-\frac{5}{3}\right \}$

$=\left \{ 33-\frac{35}{3} \right \} = \left \{ \frac{99-35}{3} \right \}$

$= \frac{64}{3}$

Question:4 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^\frac{\pi}{4}\sin 2x dx$

Answer:

Given integral: $\int_0^\frac{\pi}{4}\sin 2x dx$

Consider the integral $\int \sin 2x dx$

$\int \sin 2x dx = \frac{-\cos 2x }{2}$

So, we have the function of $x$ , $f(x) = \frac{-\cos 2x }{2}$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4})-f(0)$

$= \frac{-\cos 2(\frac{\pi}{4})}{2} + \frac{\cos 0}{2}$

$=\frac{1}{2} - 0$

$= \frac{1}{2}$

Question:5 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^\frac{\pi}{2}\cos 2x dx$

Answer:

Given integral: $\int_0^\frac{\pi}{2}\cos 2x dx$

Consider the integral $\int \cos 2x dx$

$\int \cos 2x dx = \frac{\sin 2x }{2}$

So, we have the function of $x$ , $f(x) = \frac{\sin 2x }{2}$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{2})-f(0)$

$= \frac{1}{2}\left \{ \sin 2(\frac{\pi}{2}) - \sin 0 \right \}$

$= \frac{1}{2}\left \{ 0 - 0 \right \} = 0$

Question:6 Evaluate the definite integrals in Exercises 1 to 20.

$\int_4^5 e^x dx$

Answer:

Given integral: $\int_4^5 e^x dx$

Consider the integral $\int e^x dx$

$\int e^x dx = e^x$

So, we have the function of $x$ , $f(x) = e^x$

Now, by Second fundamental theorem of calculus, we have

$I = f(5) -f(4)$

$= e^5 -e^4$

$= e^4(e-1)$

Question:7 Evaluate the definite integrals in Exercises 1 to 20.

$\int^\frac{\pi}{4}_0 \tan x dx$

Answer:

Given integral: $\int^\frac{\pi}{4}_0 \tan x dx$

Consider the integral $\int \tan x dx$

$\int \tan x dx = -\log|\cos x |$

So, we have the function of $x$ , $f(x) = -\log|\cos x |$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4}) -f(0)$

$= -\log\left | \cos \frac{\pi}{4} \right | +\log|\cos 0|$

$= -\log\left | \cos \frac{1}{\sqrt2} \right | +\log|1|$

$= -\log\left | \frac{1}{\sqrt2} \right | + 0 = -\log (2)^{-\frac{1}{2}}$

$= \frac{1}{2}\log (2)$

Question:8 Evaluate the definite integrals in Exercises 1 to 20.

$\int_\frac{\pi}{6}^\frac{\pi}{4}\textup{cosec}xdx$

Answer:

Given integral: $\int_\frac{\pi}{6}^\frac{\pi}{4}\textup{cosec}xdx$

Consider the integral $\int\textup{cosec}xdx$

$\int\textup{cosec}xdx = \log|cosec x -\cot x |$

So, we have the function of $x$ , $f(x) =\log|cosec x -\cot x |$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4}) -f(\frac{\pi}{6})$

$= \log|cosec \frac{\pi}{4} -\cot \frac{\pi}{4} | - \log|cosec \frac{\pi}{6} -\cot \frac{\pi}{6} |$

$= \log|\sqrt2 -1 | - \log|2 -\sqrt3 |$

$= \log \left ( \frac{\sqrt2 -1}{2-\sqrt3} \right )$

Question:9 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1\frac{dx}{\sqrt{1-x^2}}$

Answer:

Given integral: $\int_0^1\frac{dx}{\sqrt{1-x^2}}$

Consider the integral $\int \frac{dx}{\sqrt{1-x^2}}$

$\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x$

So, we have the function of $x$ , $f(x) = \sin^{-1}x$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \sin^{-1}(1) -\sin^{-1}(0)$

$= \frac{\pi}{2} - 0$

$= \frac{\pi}{2}$

Question:10 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1\frac{dx}{1 + x^2}$

Answer:

Given integral: $\int_0^1\frac{dx}{1 + x^2}$

Consider the integral $\int\frac{dx}{1 + x^2}$

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

So, we have the function of $x$ , $f(x) =\tan^{-1}x$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \tan^{-1}(1) -\tan^{-1}(0)$

$= \frac{\pi}{4} - 0$

$= \frac{\pi}{4}$

Question:11 Evaluate the definite integrals in Exercises 1 to 20.

$\int_2^3 \frac{dx}{x^2 -1 }$

Answer:

Given integral: $\int_2^3 \frac{dx}{x^2 -1 }$

Consider the integral $\int \frac{dx}{x^2 -1 }$

$\int \frac{dx}{x^2 -1 } = \frac{1}{2}\log\left | \frac{x-1}{x+1} \right |$

So, we have the function of $x$ , $f(x) =\frac{1}{2}\log\left | \frac{x-1}{x+1} \right |$

Now, by Second fundamental theorem of calculus, we have

$I = f(3) -f(2)$

$= \frac{1}{2}\left \{ \log\left | \frac{3-1}{3+1} \right | - \log\left | \frac{2-1}{2+1} \right | \right \}$

$= \frac{1}{2}\left \{ \log\left | \frac{2}{4} \right | -\log\left | \frac{1}{3} \right | \right \}$

$= \frac{1}{2}\left \{ \log \frac{1}{2} -\log \frac{1}{3} \right \} = \frac{1}{2}\log\frac{3}{2}$

Question:12 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^\frac{\pi}{2}\cos^2 x dx$

Answer:

Given integral: $\int_0^\frac{\pi}{2}\cos^2 x dx$

Consider the integral $\int \cos^2 x dx$

$\int \cos^2 x dx = \int \frac{1+\cos 2x}{2} dx = \frac{x}{2}+\frac{\sin 2x }{4}$

$= \frac{1}{2}\left ( x+\frac{\sin 2x}{2} \right )$

So, we have the function of $x$ , $f(x) =\frac{1}{2}\left ( x+\frac{\sin 2x}{2} \right )$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{2}) -f(0)$

$= \frac{1}{2}\left \{ \left ( \frac{\pi}{2}-\frac{\sin \pi}{2} \right ) -\left ( 0+\frac{\sin 0}{2} \right ) \right \}$

$= \frac{1}{2}\left \{ \frac{\pi}{2}+0-0-0 \right \}$

$= \frac{\pi}{4}$

Question:13 Evaluate the definite integrals in Exercises 1 to 20.

$\int_2^3\frac{xdx}{x^2+1}$

Answer:

Given integral: $\int_2^3\frac{xdx}{x^2+1}$

Consider the integral $\int \frac{xdx}{x^2+1}$

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

So, we have the function of $x$ , $f(x) =\frac{1}{2}\log(1+x^2)$

Now, by Second fundamental theorem of calculus, we have

$I = f(3) -f(2)$

$= \frac{1}{2}\left \{ \log(1+(3)^2)-\log(1+(2)^2) \right \}$

$= \frac{1}{2}\left \{ \log(10)-\log(5) \right \} = \frac{1}{2}\log\left ( \frac{10}{5} \right ) = \frac{1}{2}\log2$

Question:14 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1\frac{2x+3}{5x^2+1}dx$

Answer:

Given integral: $\int_0^1\frac{2x+3}{5x^2+1}dx$

Consider the integral $\int \frac{2x+3}{5x^2+1}dx$

Multiplying by 5 both in numerator and denominator:

$\int \frac{2x+3}{5x^2+1}dx = \frac{1}{5}\int \frac{5(2x+3)}{5x^2+1}dx$

$=\frac{1}{5}\int \frac{10x+15}{5x^2+1}dx$

$= \frac{1}{5} \int \frac{10x}{5x^2+1} dx +3\int \frac{1}{5x^2+1} dx$

$= \frac{1}{5}\int \frac{10x}{5x^2+1}+3\int \frac{1}{5\left ( x^2+\frac{1}{5} \right )}dx$

$= \frac{1}{5}\log(5x^2+1) +\frac{3}{5}\times \frac{1}{\frac{1}{\sqrt5}} \tan^{-1}\frac{x}{\frac{1}{\sqrt5}}$

$= \frac{1}{5}\log(5x^2+1) +\frac{3}{\sqrt5}\tan^{-1}(\sqrt5 x )$

So, we have the function of $x$ , $f(x) = \frac{1}{5}\log(5x^2+1) +\frac{3}{\sqrt5}\tan^{-1}(\sqrt5 x )$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \left \{ \frac{1}{5}\log(1+5)+\frac{3}{\sqrt5}\tan^{-1}(\sqrt5) \right \} - \left \{ \frac{1}{5}\log(1)+\frac{3}{\sqrt5}\tan^{-1}(0) \right \}$

$= \frac{1}{5}\log 6 +\frac{3}{\sqrt 5}\tan^{-1}{\sqrt5}$

Question:15 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1xe^{x^2}dx$

Answer:

Given integral: $\int_0^1xe^{x^2}dx$

Consider the integral $\int xe^{x^2}dx$

Putting $x^2 = t$ which gives, $2xdx =dt$

As, $x\rightarrow0 ,t \rightarrow0$ and as $x\rightarrow1 ,t \rightarrow1$ .

So, we have now:

$\therefore I = \frac{1}{2}\int_0^1 e^t dt$

$= \frac{1}{2}\int e^t dt = \frac{1}{2} e^t$

So, we have the function of $x$ , $f(x) = \frac{1}{2} e^t$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \frac{1}{2}e^1 -\frac{1}{2}e^0 = \frac{1}{2}(e-1)$

Question:16 Evaluate the definite integrals in Exercises 1 to 20.

$\int_1^2\frac{5x^2}{x^2 + 4x +3}$

Answer:

Given integral: $I = \int_1^2\frac{5x^2}{x^2 + 4x +3}$

So, we can rewrite the integral as;

$I = \int_1^2 \frac{5x^2}{x^2 + 4x +3}= \int_1^2 \left ( 5 - \frac{20x+15}{x^2 + 4x +3} \right ) dx$

$= \int_1^2 5 dx - \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

$= [5x]_1^2 - \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

$I = 5-I_1$ where $I = \int_1^2 \frac{20x+15}{x^2+4x+3}dx$ . ................(1)

Now, consider $I = \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

Take numerator $20x+15 = A \frac{d}{dx}\left ( x^2+4x+3 \right )+B$

$= 2A x+(4A+B)$

We now equate the coefficients of x and constant term, we get

A=10 and B=-25

$\Rightarrow I_1 = 10\int_1^2 \frac{2x+4}{x^2+4x+3}dx -25\int_1^2 \frac{dx}{x^2+4x+3}$

Now take denominator $x^2+4x+3 = t$

Then we have $(2x+4)dx =dt$

$\Rightarrow I_{1} =10\int \frac{dt}{t} -25\int \frac{dx}{(x+2)^2-1^2}$

$= 10\log t -25\left [ \frac{1}{2}\log\left ( \frac{x+2-1}{x+2+1} \right ) \right ]$

$=[10\log(x^2+4x+3)]_1^2 -25 \left [ \frac{1}{2}\log\left ( \frac{x+1}{x+3} \right ) \right ]_1^2$

$= \left [ 10\log15 -10\log 8 \right ] -25 \left [ \frac{1}{2}\log\frac{3}{5} -\frac{1}{2}\log\frac{2}{4} \right ]$

$= \left [ 10\log5 +10\log3 -10\log4-10\log2 \right ] -\frac{25}{2}\left [ \log3 -\log5-\log2+\log4 \right ]$ $= \left ( 10+\frac{25}{2} \right )\log5 + \left ( -10-\frac{25}{2} \right )\log 4 + \left ( 10-\frac{25}{2} \right )\log 3 + \left ( -10+\frac{25}{2} \right )\log 2$ $= \frac{45}{2}\log5 -\frac{45}{2}\log4 - \frac{5}{2}\log3 +\frac{5}{2}\log2$

$= \frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log \frac{3}{2}$

Then substituting the value of $I_{1}$ in equation (1), we get

$I= 5 -\left ( \frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log\frac{3}{2} \right )$

$= 5 -\frac{5}{2}\left ( 9\log\frac{5}{4}-\log\frac{3}{2} \right )$

Question:17 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^\frac{\pi}{4}(2\sec^2x + x^3 + 2)dx$

Answer:

Given integral: $\int_0^\frac{\pi}{4}(2\sec^2x + x^3 + 2)dx$

Consider the integral $\int (2\sec^2x + x^3 + 2)dx$

$\int (2\sec^2x + x^3 + 2)dx = 2\tan x +\frac{x^4}{4}+2x$

So, we have the function of $x$ , $f(x) = 2\tan x +\frac{x^4}{4}+2x$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4}) -f(0)$

$= \left \{ \left ( 2\tan\frac{\pi}{4}+\frac{1}{4}\left ( \frac{\pi}{4} \right )^4+2\frac{\pi}{4} \right ) - \left ( 2\tan 0 +0 +0 \right ) \right \}$

$=2\tan\frac{\pi}{4} +\frac{\pi^4}{4^5} +\frac{\pi}{2}$

$2+\frac{\pi}{2}+\frac{\pi^4}{1024}$

Question:18 Evaluate the definite integrals in Exercises 1 to 20.

$\int^\pi_0(\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx$

Answer:

Given integral: $\int^\pi_0(\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx$

Consider the integral $\int (\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx$

can be rewritten as: $-\int (\cos^2\frac{x}{2} - \sin^2\frac{x}{2})dx = -\int_0^{\pi} \cos x dx$

$= \sin x$

So, we have the function of $x$ , $f(x) =\sin x$

Now, by Second fundamental theorem of calculus, we have

$I = f(\pi) - f(0)$

$\Rightarrow \sin \pi - \sin 0 = 0-0 =0$

Question:19 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^2\frac{6x+3}{x^2+ 4}$

Answer:

Given integral: $\int_0^2\frac{6x+3}{x^2+ 4}$

Consider the integral $\int \frac{6x+3}{x^2+ 4}$

can be rewritten as: $\int \frac{6x+3}{x^2+ 4} = 3\int \frac{2x+1}{x^2+4}dx$

$= 3\int \frac{2x}{x^2+4}dx +3\int \frac{1}{x^2+4}dx$

$= 3\log (x^2+4) +\frac{3}{2}\tan^{-1}\frac{x}{2}$

So, we have the function of $x$ , $f(x) =3\log (x^2+4) +\frac{3}{2}\tan^{-1}\frac{x}{2}$

Now, by Second fundamental theorem of calculus, we have

$I = f(2) - f(0)$

$= \left \{ 3\log(2^2+4)+\frac{3}{2}\tan^{-1}\left ( \frac{2}{2} \right ) \right \}- \left \{ 3\log(0+4)+\frac{3}{2}\tan^{-1}\left ( \frac{0}{2} \right ) \right \}$ $=3\log 8 +\frac{3}{2}\tan^{-1}1 -3\log 4 -\frac{3}{2}\tan^{-1} 0$

$=3\log 8 +\frac{3}{2}\times\frac{\pi}{4} -3\log 4 -0$

$=3\log \frac{8}{4} +\frac{3\pi}{8}$

or we have $=3\log 2 +\frac{3\pi}{8}$

Question:20 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1(xe^x + sin\frac{\pi x}{4})dx$

Answer:

Given integral: $\int_0^1(xe^x + sin\frac{\pi x}{4})dx$

Consider the integral $\int (xe^x + sin\frac{\pi x}{4})dx$

can be rewritten as: $x\int e^x dx - \int \left \{ \left ( \frac{d}{dx}x \right )\int e^x dx \right \}dx +\left \{ \frac{-\cos \frac{\pi x}{4}}{\frac{\pi}{4}} \right \}$

$= xe^x -\int e^x dx -\frac{4\pi}{\pi} \cos \frac{x}{4}$

$= xe^x -e^x -\frac{4\pi}{\pi} \cos \frac{x}{4}$

So, we have the function of $x$ , $f(x) = xe^x -e^x -\frac{4\pi}{\pi} \cos \frac{x}{4}$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) - f(0)$

$= \left (1.e^t-e^t - \frac{4}{\pi}\cos \frac{\pi}{4} \right ) - \left ( 0.e^0 -e^0 -\frac{4}{\pi}\cos 0 \right )$

$= e-e -\frac{4}{\pi}\left ( \frac{1}{\sqrt2} \right )+1+\frac{4}{\pi}$

Question:21 Choose the correct answer in Exercises 20 and 21.

$\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2}$

(A) $\frac{\pi}{3}$

(B) $\frac{2\pi}{3}$

(D) $\frac{\pi}{12}$

Answer:

Given definite integral $\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2}$

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

we have then the function of x, as $f(x) = \tan^{-1}x$

By applying the second fundamental theorem of calculus, we will get

$\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2} = f(\sqrt3) - f(1)$

$= \tan^{-1}\sqrt{3} - \tan^{-1}1$

$=\frac{\pi}{3} - \frac{\pi}{4}$

$= \frac{\pi}{12}$

Therefore the correct answer is D.

Question:22 Choose the correct answer in Exercises 21 and 22.

$\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2}$ equals

(A) $\frac{\pi}{6}$

(B) $\frac{\pi}{12}$

(D) $\frac{\pi}{4}$

Answer:

Given definite integral $\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2}$

Consider $\int \frac{dx}{4+ 9x^2} = \int \frac{dx}{2^2+(3x)^2}$

Now, putting $3x = t$

we get, $3dx=dt$

Therefore we have, $\int \frac{dx}{2^2+(3x)^2} = \frac{1}{3}\int \frac{dt}{2^2+t^2}$

$= \frac{1}{3}\left ( \frac{1}{2}\tan^{-1}\frac{t}{2} \right ) = \frac{1}{6}\tan^{-1}\left ( \frac{3x}{2} \right )$

we have the function of x , as $f(x) =\frac{1}{6}\tan^{-1}\left ( \frac{3x}{2} \right )$

So, by applying the second fundamental theorem of calculus, we get

$\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2} = f(\frac{2}{3}) - f(0)$

$= \frac{1}{6}\tan^{-1}\left ( \frac{3}{2}.\frac{2}{3} \right ) -\frac{1}{6}\tan^{-1}0$

$= \frac{1}{6}\tan^{-1}1 - 0$

$= \frac{1}{6}\times \frac{\pi}{4} = \frac{\pi}{24}$

Therefore the correct answer is C.

Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.9

The Class 12th Maths chapter 7 exercise is a one stop solution for clearing concepts as well as practicing questions.
Practicing exercise 7.9 Class 12 Maths can be of great help to the students aspiring to score well in the examination.
These Class 12 Maths chapter 7 exercise 7.9 solutions can be asked directly in Board examination which can be verified from looking Previous year questions.
Along with NCERT solutions for Class 12 Maths chapter 7 exercise 7.9 one must also go through the previous year questions of the same topics.

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Comprehensive Coverage: The solutions encompass all the topics covered in ex 7.9 class 12, ensuring a thorough understanding of the concepts.
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Accuracy and Clarity: Solutions for class 12 ex 7.9 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this 12th class maths exercise 7.9 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
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Frequently Asked Questions (FAQs)

1. What do you mean by definite integrals ?

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

2. What is the importance of Exercise 7.9 Class 12 Maths?

Direct questions from this exercise are asked in the Board examination. Hence this exercise cannot be avoided at any cost.

3. What is the use of learning applications of Integrals ?

After learning Application of integrals, one can easily find the quantities of area, volume, displacement etc.

4. What is the difficulty level of questions of Exercise 7.9 Class 12 Maths?

Questions are moderate to difficult but regular practice can help get through the difficulty.

5. Which topics are mainly discussed in the Exercise 7.9 Class 12 Maths?

Exercise 7.9 Class 12 Maths discusses maily evaluation of definite integrals.

6. How many questions are there in Exercise 7.9 Class 12 Maths?

Exercise 7.9 Class 12 Maths has 22 questions.

Also Read

NCERT Class 12 Maths Notes - Download Chapter wise Free PDFs

NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

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NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

NCERT Class 12 Physics Chapter 6 Notes Electromagnetic Induction - Download PDF

NCERT Class 12 Physics Chapter 5 Notes Magnetism and Matter - Download PDF

NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

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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:

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.
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.
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.
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.
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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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

NCERT Solutions for Exercise 7.9 Class 12 Maths Chapter 7 - Integrals

NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.9

VMC VIQ Scholarship Test

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Access NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.9

Question:1 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:2 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:3 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:4 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:5 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:6 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:7 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:8 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:9 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:10 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:11 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:12 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:13 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:14 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:15 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:16 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:17 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:18 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:19 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:20 Evaluate the definite integrals in Exercises 1 to 20.

Answer:

Question:21 Choose the correct answer in Exercises 20 and 21.

Answer:

Question:22 Choose the correct answer in Exercises 21 and 22.

Answer:

More About NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.9

Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.9

Key Features Of NCERT Solutions for Exercise 7.9 Class 12 Maths Chapter 7

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