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

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}

(C) \frac{\pi}{6}

(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}

(C) \frac{\pi}{24}

(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.

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

The NCERT Class 12 Maths chapter Integrals is one of the most important chapters of Class 12 Maths NCERT syllabus. If we talk about exercise 7.9 Class 12 Maths, it holds good weightage in the Board examination. NCERT Solutions for Class 12 Maths chapter 7 exercise 7.9 can be learnt easily if practice of some questions is done on a regular basis.

Also Read| Integrals Class 12 Notes

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

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 7.9 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 7.9, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • 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.
  • Inclusive Approach: Solutions for ex 7.9 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 7.9 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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Frequently Asked Question (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.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
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.

You can use them people also used problem

Read Complete Answer
Divyan 06 August,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,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

To apply, download the Medhavi App from the Google Play Store, sign up, and read the detailed notification about the scholarship exam. Complete the registration within the app, take the exam from home using the app, and receive your results within two days. Following this, upload the necessary documents and bank account details for verification. Upon successful verification, the scholarship amount will be directly transferred to your bank account.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Read Complete Answer
Yuvan 01 September,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
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

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 help you.

Good luck with your studies!

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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