Skip to main content

Class 12 Maths Chapter 7 Integrals (DEFINITE INTEGRALS) EX 7.8 TO EX 7.11

Exercise 7.8

Evaluate the following definite integral as limit of sums.


 Question 1.
\int _{ a }^{ b }{ x\quad dx }
Solution:
on comparing
\int _{ a }^{ b }{ x\quad dx } \quad with\quad \int _{ a }^{ b }{ f(x)dx }
we have
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 1

Question 2
\int _{ 0 }^{ 5 }{ (x+1)dx }
Solution:
on comparing
\int _{ 0 }^{ 5 }{ (x+1)dx } \quad with\quad \int _{ 0 }^{ 5 }{ f(x)dx }
we have f(x) = x+1, a = 0, b = 5
and nh = b-a = 5-0 = 5
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 2

Question 3.
\int _{ 2 }^{ 3 }{ { x }^{ 2 } } dx
Solution:
compare
\int _{ 2 }^{ 3 }{ { x }^{ 2 } } dx\quad with\quad \int _{ a }^{ b }{ f({ x }) } dx
we have
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 3

Question 4.
\int _{ 1 }^{ 4 }{ ({ x }^{ 2 }-x) } dx
Solution:
compare
\int _{ 1 }^{ 4 }{ ({ x }^{ 2 }-x) } dx\quad with\quad \int _{ a }^{ b }{ f({ x }) } dx
we have f(x) = x²-x and a = 1, b = 4
vedantu class 12 maths Chapter 7 Integrals 4

Question 5.
\int _{ -1 }^{ 1 }{ { e }^{ x } } dx\quad
Solution:
compare
\int _{ -1 }^{ 1 }{ { e }^{ x } } dx\quad with\quad \int _{ a }^{ b }{ f({ x }) } dx
we have
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 5
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 5.1

Question 6.
\int _{ 0 }^{ 4 }{ { (x+e }^{ 2x }) } dx\quad
Solution:
let f(x) = x + e2x,
a = 0, b = 4
and nh = b – a = 4 – 0 = 4

Exercise 7.9

Evaluate the definite integrals in Exercise 1 to 20.

Question 1.
\int _{ -1 }^{ 1 }{ { (x+1 }) } dx\quad
Solution:
{ =\left[ \frac { { x }^{ 2 } }{ 2 } +x \right] }_{ -1 }^{ 1 }=\frac { 1 }{ 2 } (1-1)+(1+1)\quad =2

Question 2.
\int _{ 2 }^{ 3 }{ \frac { 1 }{ x } dx }
Solution:
={ \left[ log\quad x \right] }_{ 2 }^{ 3 }\quad =log3-log2\quad =log\frac { 3 }{ 2 }

Question 3.
\int _{ 1 }^{ 2 }{ \left( { 4x }^{ 3 }-{ 5x }^{ 2 }+6x+9 \right) dx }
Solution:
={ \left[ \frac { { 4x }^{ 4 } }{ 4 } -\frac { { 5x }^{ 3 } }{ 3 } +\frac { { 6x }^{ 2 } }{ 2 } +9x \right] }_{ 1 }^{ 2 }
={ \left[ { x }^{ 4 }-\frac { 5 }{ 3 } { x }^{ 3 }+{ 3x }^{ 2 }+9x \right] }_{ 1 }^{ 2 }\quad =\frac { 64 }{ 3 }

Question 4.
\int _{ 0 }^{ \frac { \pi }{ 4 } }{ sin2x\quad dx }
Solution:
={ \left[ -\frac { 1 }{ 2 } cos2x \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\quad =\frac { 1 }{ 2 }

Question 5.
\int _{ 0 }^{ \frac { \pi }{ 2 } }{ cos2x\quad dx }
Solution:
={ \left[ \frac { 1 }{ 2 } sin2x \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\quad =0

Question 6.
\int _{ 4 }^{ 5 }{ { e }^{ x }dx }
Solution:
={ \left[ { e }^{ x } \right] }_{ 4 }^{ 5 }\quad ={ e }^{ 5 }-{ e }^{ 4 }

Question 7.
\int _{ 0 }^{ \frac { \pi }{ 4 } }{ tanx\quad dx }
Solution:
={ \left[ log\quad secx \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\quad =\frac { 1 }{ 2 } log2

Question 8.
\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 4 } }{ cosec\quad xdx }
Solution:
=log{ \left( cosecx-cotx \right) }_{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 4 } }
=log(\sqrt { 2 } -1)-log(2-\sqrt { 3 } )\quad =log\left( \frac { \sqrt { 2 } -1 }{ 2-\sqrt { 3 } } \right)

Question 9.
\int _{ 0 }^{ 1 }{ \frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } }
Solution:
={ sin }^{ -1 }(1)-{ sin }^{ -1 }(0)\quad =\frac { \pi }{ 2 }

Question 10.
\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+{ x }^{ 2 } } }
Solution:
={ \left[ { tan }^{ -1 }x \right] }_{ 0 }^{ 1 }\quad ={ tan }^{ -1 }(1)-{ ta }n^{ -1 }(0)\quad =\frac { \pi }{ 4 }

Question 11.
\int _{ 2 }^{ 3 }{ \frac { dx }{ { x }^{ 2 }-1 } }
Solution:
={ \left[ \frac { 1 }{ 2 } log\left( \frac { x-1 }{ x+1 } \right) \right] }_{ 2 }^{ 3 }\quad =\frac { 1 }{ 2 } log\frac { 3 }{ 2 }

Question 12.
\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 2 } } xdx
Solution:
=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \frac { 1+cos2x }{ 2 } } } dx=\frac { 1 }{ 2 } { \left[ x+\frac { sin2x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=\frac { \pi }{ 4 }

Question 13.
\int _{ 2 }^{ 3 }{ \frac { x }{ { x }^{ 2 }+1 } } dx
Solution:
=\frac { 1 }{ 2 } \int _{ 2 }^{ 3 }{ \frac { 2x }{ { x }^{ 2 }+1 } } dx\quad =\frac { 1 }{ 2 } { \left[ log\left( { x }^{ 2 }+1 \right) \right] }_{ 2 }^{ 3 }\quad =\frac { 1 }{ 2 } log2

Question 14.
\int _{ 0 }^{ 1 }{ \frac { 2x+3 }{ { 5x }^{ 2 }+1 } dx }
Solution:
=\frac { 1 }{ 5 } \int _{ 0 }^{ 1 }{ \frac { 10x }{ { 5x }^{ 2 }+1 } dx } +\frac { 3 }{ 5 } \int _{ 0 }^{ 1 }{ \frac { dx }{ { { x }^{ 2 }+\left[ \frac { 1 }{ \sqrt { 5 } } \right] }^{ 2 } } }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 14

Question 15.
\int _{ 0 }^{ 1 }{ { xe }^{ { x }^{ 2 } }dx }
Solution:
let x² = t ⇒ 2xdx = dt
when x = 0, t = 0 & when x = 1,t = 1
\therefore I=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ { e }^{ t }dt } \quad =\frac { 1 }{ 2 } { \left( { e }^{ t } \right) }_{ 0 }^{ 1 }\quad =\frac { 1 }{ 2 } [e-1]

Question 16.
\int _{ 1 }^{ 2 }{ \frac { { 5x }^{ 2 } }{ { x }^{ 2 }+4x+3 } dx }
Solution:
\int _{ 1 }^{ 2 }{ \left( 5-\frac { 20x+15 }{ { x }^{ 2 }+4x+3 } \right) dx }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 16

Question 17.
\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( { 2sec }^{ 2 }x+{ x }^{ 3 }+2 \right) dx }
Solution:
={ \left[ 2tanx+\frac { { x }^{ 4 } }{ 4 } +2x \right] }_{ 0 }^{ \frac { \pi }{ 4 } }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 17

Question 18.
\int _{ 0 }^{ \pi }{ \left( { sin }^{ 2 }\frac { x }{ 2 } -{ cos }^{ 2 }\frac { x }{ 2 } \right) } dx
Solution:
=-\int _{ 0 }^{ \pi }{ cosx } dx\quad =-{ \left[ sinx \right] }_{ 0 }^{ \pi }-(0-0)\quad =0

Question 19.
\int _{ 0 }^{ 2 }{ \frac { 6x+3 }{ { x }^{ 2 }+4 } } dx
Solution:
=\int _{ 0 }^{ 2 }{ \frac { 6x }{ { x }^{ 2 }+4 } } dx+\int _{ 0 }^{ 2 }{ \frac { 3 }{ { x }^{ 2 }+4 } dx }
byjus class 12 maths Chapter 7 Integrals 19

Question 20.
\int _{ 0 }^{ 1 }{ \left( { xe }^{ x }+sin\frac { \pi x }{ 4 } \right) dx }
Solution:
=\int _{ 0 }^{ 1 }{ { xe }^{ x }dx } +\int _{ 0 }^{ 1 }{ sin\frac { \pi x }{ 4 } } dx
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 20

Question 21.
\int _{ 1 }^{ \sqrt { 3 } }{ \frac { dx }{ { 1+x }^{ 2 } } \quad equals }
(a) \frac { \pi }{ 3 }
(b) \frac { 2\pi }{ 3 }
(c) \frac { \pi }{ 6 }
(d) \frac { \pi }{ 12 }
Solution:
(d) \int _{ 1 }^{ \sqrt { 3 } }{ \frac { dx }{ { 1+x }^{ 2 } } } \quad ={ \left[ { tan }^{ -1 }x \right] }_{ 1 }^{ \sqrt { 3 } }\quad =\frac { \pi }{ 3 } -\frac { \pi }{ 4 } \quad =\frac { \pi }{ 12 }

Question 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 }
Solution:
(c) \int _{ 0 }^{ \frac { 2 }{ 3 } }{ \frac { dx }{ 4+{ 9x }^{ 2 } } } \quad =\frac { 1 }{ 9 } \int _{ 0 }^{ \frac { 2 }{ 3 } }{ \frac { dx }{ { \left( \frac { 2 }{ 3 } \right) }^{ 2 }+{ x }^{ 2 } } }
=\frac { 1 }{ 6 } { \left[ { tan }^{ -1 }\left( \frac { 3x }{ 2 } \right) \right] }_{ 0 }^{ \frac { 2 }{ 3 } }\quad =\frac { 1 }{ 6 } \times \frac { \pi }{ 4 } \quad =\frac { \pi }{ 24 }

Exercise 7.10

Evaluate the integrals in Exercises 1 to 8 using substitution.

Question 1.
\int _{ 0 }^{ 1 }{ \frac { x }{ { x }^{ 2 }+1 } } dx=I
Solution:
Let x² + 1 = t
⇒2xdx = dt
when x = 0, t = 1 and when x = 1, t = 2
\therefore I=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { dt }{ t } } ={ \left[ \frac { 1 }{ 2logt } \right] }_{ 1 }^{ 2 }\quad =\frac { 1 }{ 2 } log2

Question 2.
\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { sin\phi } { cos }^{ 5 }\phi d\phi =I }
Solution:
I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { sin\phi } { (1-{ sin }^{ 2 }) }^{ 2 }cos\phi d\phi }
put sinφ = t,so that cosφdφ = dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 2

Question 3.
\int _{ 0 }^{ 1 }{ { sin }^{ -1 } } \left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) dx=I
Solution:
let x = tanθ =>dx = sec²θ dθ
when x = 0 => θ = 0
and when x = 1 => \theta \frac { \pi }{ 4 }
\frac { 1 }{ 2 }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 3

Question 4.
\int _{ 0 }^{ 2 }{ x\sqrt { x+2 } } dx=I(say)(put\quad x+2={ t }^{ 2 })
Solution:
let x+2 = t =>dx = dt
when x = 0,t = 2 and when x = 2, t = 4
tiwari academy class 12 maths Chapter 7 Integrals 4

Question 5.
\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx\quad dx }{ 1+{ cos }^{ 2 }x } =I }
Solution:
put cosx = t
so that -sinx dx = dt
when x = 0, t = 1; when x=\frac { \pi }{ 2 }, t = 0
\therefore I=\int _{ 1 }^{ 0 }{ \frac { -dt }{ 1+{ t }^{ 2 } } =-{ \left[ { tan }^{ -1 }t \right] }_{ 1 }^{ 0 } } =\frac { \pi }{ 4 }

Question 6.
\int _{ 0 }^{ 2 }{ \frac { dx }{ x+4-{ x }^{ 2 } } =I }
Solution:
\int _{ 0 }^{ 2 }{ \frac { dx }{ x+4-{ x }^{ 2 } } =I }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 6

Question 7.
\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+5 } =I }
Solution:
I=\int _{ -1 }^{ 1 }{ \frac { dx }{ { (x+1) }^{ 2 }+{ 2 }^{ 2 } } } =\frac { 1 }{ 2 } { \left[ { tan }^{ -1 }\frac { x+1 }{ 2 } \right] }_{ -1 }^{ 1 }\quad =\frac { \pi }{ 8 }

Question 8.
\int _{ 1 }^{ 2 }{ \left[ \frac { 1 }{ x } -\frac { 1 }{ { 2x }^{ 2 } } \right] { e }^{ 2x }dx } =I
Solution:
let 2x = t ⇒ 2dx = dt
when x = 1, t = 2 and when x = 2, t = 4
I=\int _{ 2 }^{ 4 }{ e } ^{ t }\left( \frac { 1 }{ t } -\frac { 1 }{ { t }^{ 2 } } \right) dt\quad ={ e }^{ t }{ \left[ \frac { 1 }{ t } \right] }_{ 2 }^{ 4 }\quad =\frac { e^{ 2 } }{ 2 } \left[ \frac { { e }^{ 2 } }{ 2 } -1 \right]

Choose the correct answer in Exercises 9 and 10

Question 9.
The value of integral \int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { { { (x-x }^{ 3 }) }^{ \frac { 1 }{ 3 } } }{ { x }^{ 4 } } dx } is
(a) 6
(b) 0
(c) 3
(d) 4
Solution:
(a) let I = \int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { { { (x-x }^{ 3 }) }^{ \frac { 1 }{ 3 } } }{ { x }^{ 4 } } dx } \quad =\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { { x }^{ \frac { 1 }{ 3 } }(1-{ x }^{ 2 })^{ \frac { 1 }{ 3 } } }{ { x }^{ 4 } } dx }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 9

Question 10.
If\quad f(x)=\int _{ 0 }^{ x }{ tsint,\quad then\quad { f }^{ \prime }(x)\quad is }
(a) cosx+xsinx
(b) xsinx
(c) xcosx
(d) sinx+xcosx
Solution:
(b) f(x)=\int _{ 0 }^{ x }{ tsint\quad dt }
=t(-cost)-\int { 1{ \left[ (-cost)dt \right] }_{ 0 }^{ x } }
=-x cox+sinx

Exercise 7.11

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Question 1.
\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 2 }x\quad dx } =I
Solution:
I=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ (1+cos2x)dx } =\frac { 1 }{ 2 } { \left[ x+\frac { sin2x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\quad =\frac { \pi }{ 4 }

Question 2.
\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { sinx } }{ \sqrt { sinx } +\sqrt { cosx } } dx }
Solution:
let I = \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { sinx } }{ \sqrt { sinx } +\sqrt { cosx } } dx }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 2

Question 3.
\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sin }^{ \frac { 3 }{ 2 } }xdx }{ { sin }^{ \frac { 3 }{ 2 } }x+{ cos }^{ \frac { 3 }{ 2 } }dx } dx }
Solution:
let I = \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sin }^{ \frac { 3 }{ 2 } }xdx }{ { sin }^{ \frac { 3 }{ 2 } }x+{ cos }^{ \frac { 3 }{ 2 } }dx } dx }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 3

Question 4.
\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { cos }^{ 5 }xdx }{ { sin }^{ 5 }x+{ cos }^{ 5 }x } }
Solution:
let I = \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { cos }^{ 5 }xdx }{ { sin }^{ 5 }x+{ cos }^{ 5 }x } }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 4

Question 5.
\int _{ -5 }^{ 5 }{ \left| x+2 \right| dx=I }
Solution:
I=\int _{ -5 }^{ 5 }{ \left| x+2 \right| dx+\int _{ -2 }^{ 5 }{ \left| x+2 \right| dx } }
at x = – 5, x + 2 < 0; at x = – 2, x + 2 = 0; at x = 5, x + 2>0;x + 2<0, x + 2 = 0, x + 2>0
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 5

Question 6.
\int _{ 2 }^{ 8 }{ |x-5|dx } =I
Solution:
\int _{ 2 }^{ 8 }{ |x-5|dx } =I
vedantu class 12 maths Chapter 7 Integrals 6

Question 7.
\int _{ 0 }^{ 1 }{ x(1-x)^{ n }dx } =I
Solution:
\int _{ 0 }^{ 1 }{ x(1-x)^{ n }dx } =I
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 7

Question 8.
\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tanx)dx }
Solution:
let I = \int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tanx)dx }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 8

Question 9.
\int _{ 0 }^{ 2 }{ x\sqrt { 2-x } dx=I }
Solution:
let 2-x = t
⇒ – dx = dt
when x = 0, t = 2 and when x = 2,t = 0
\frac { 1 }{ 2 }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 9.

Question 10.
\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2logsinx-logsin2x \right) dx=I }
Solution:
\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2logsinx-logsin2x \right) dx=I }
vedantu class 12 maths Chapter 7 Integrals 10

Question 11.
\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { sin }^{ 2 } } xdx
Solution:
Let f(x) = sin² x
f(-x) = sin² x = f(x)
∴ f(x) is an even function
\therefore \int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { sin }^{ 2 } } xdx\quad =2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left[ \frac { 1-cos2x }{ 2 } \right] dx }
={ \left[ x-\frac { sin2x }{ x } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\therefore I=\frac { \pi }{ 2 }

Question 12.
\int _{ 0 }^{ \pi }{ \frac { xdx }{ 1+sinx } }
Solution:
let I = \int _{ 0 }^{ \pi }{ \frac { xdx }{ 1+sinx } } …(i)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 12

Question 13.
\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { sin }^{ 7 } } xdx
Solution:
Let f(x) = sin7 xdx
⇒ f(-x) = -sin7 x = -f(x)
⇒ f(x) is an odd function of x
⇒ \int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { sin }^{ 7 } } xdx=0

Question 14.
\int _{ 0 }^{ 2\pi }{ { cos }^{ 5 } } xdx
Solution:
let f(x) = cos5 x
⇒ f(2π – x) = cos5 x
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 14
vedantu class 12 maths Chapter 7 Integrals 14.1

Question 15.
\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx-cosx }{ 1+sinx\quad cosx } dx }
Solution:
let I = \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx-cosx }{ 1+sinx\quad cosx } dx } …(i)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 15

Question 16.
\int _{ 0 }^{ \pi }{ log(1+cosx)dx }
Solution:
let I = \int _{ 0 }^{ \pi }{ log(1+cosx)dx }
then I = \int _{ 0 }^{ \pi }{ log[1+cos(\pi -x)]dx }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 16
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 16.1

Question 17.
\int _{ 0 }^{ a }{ \frac { \sqrt { x } }{ \sqrt { x } +\sqrt { a-x } } dx }
Solution:
let I = \int _{ 0 }^{ a }{ \frac { \sqrt { x } }{ \sqrt { x } +\sqrt { a-x } } dx } …(i)
vedantu class 12 maths Chapter 7 Integrals 17

Question 18.
\int _{ 0 }^{ 4 }{ \left| x-1 \right| dx=I }
Solution:
I=-\int _{ 0 }^{ 1 }{ (x-1)dx } +\int _{ 1 }^{ 4 }{ (x-1)dx }
=-{ \left[ \frac { { x }^{ 2 } }{ 2 } -x \right] }_{ 0 }^{ 1 }+{ \left[ \frac { { x }^{ 2 } }{ 2 } -x \right] }_{ 1 }^{ 4 }=5

Question 19.
show that 4\int _{ 0 }^{ a }{ f(x)g(x)dx } =2\int _{ 0 }^{ a }{ f(x)dx } if f and g are defined as f(x)=f(a-x) and g(x)+g(a-x)=4
Solution:
let I = \int _{ 0 }^{ a }{ f(x)g(x)dx }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 19

Question 20.
The value of \int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \left( { x }^{ 3 }+xcosx+{ tan }^{ 5 }x+1 \right) dx } is
(a) 0
(b) 2
(c) π
(d) 1
Solution:
(c) let I = \int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \left( { x }^{ 3 }+xcosx+{ tan }^{ 5 }x+1 \right) dx } is
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 20

Question 21.
The value of \int _{ 0 }^{ \frac { \pi }{ 2 } }{ log\left[ \frac { 4+3sinx }{ 4+3sinx } \right] dx } is
(a) 2
(b) \frac { 3 }{ 4 }
(c) 0
(d) -2
Solution:
let I = \int _{ 0 }^{ \frac { \pi }{ 2 } }{ log\left[ \frac { 4+3sinx }{ 4+3sinx } \right] dx }
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 21
NCERT Solutions for Class 12 Maths Chapter 7 Integrals 21.1


Comments

Post a Comment

Popular posts from this blog

GRAMMAR:-Preposition Exercises with Answers

A preposition is a word placed before a Noun or Pronoun to show its relation to some other word in the sentence. Some examples of prepositions are words like "in", "at", "on", "of", and "to". Prepositions in English are highly idiomatic. Fill the blanks with the suitable prepositions 1) The dog ran ____ the road. 2) The river flows ____ the bridge. 3) The work was done ____ haste. 4) He is afraid ____ the dog. 5) I am fond ____ music. 6) He goes ____ Sunday ____ church. 7) He died ____ his country. 8) The steam-engine was invented ____ James Watt. 9) The burglar jumped ____ the compound wall. 10) The village was destroyed ____ fire. 11) What is that ____ me? 12) It cannot be done ____ offence. 13) He spoke ____ me ____ Urdu. 14) They live ____ the same roof. 15) I have not seen him ____ Wednesday last. 16) I have known him ____ a long time. 17) The moon does not shine ____ its own light. 18) This is a matter ____ little importance. 19)...
Post a Comment
Read more

Acceleration

  acceleration  is the  rate  of change of  velocity  of an object with respect to time.  An object's acceleration is the net result of any and all  forces  acting on the object, as described by  Newton's Second Law . The  SI  unit for acceleration is  metre per second squared   (m s −2 ).  Accelerations are  vector  quantities (they have  magnitude  and  direction ) and add according to the  parallelogram law . [2] [3]  As a  vector , the calculated net  force  is equal to the product of the object's mass (a  scalar  quantity) and its acceleration. For example, when a car starts from a standstill (zero relative velocity) and travels in a straight line at increasing speeds, it is accelerating in the direction of travel. If the car turns, an acceleration occurs toward the new direction. Average acceleration An object's average acceleration over a...
Post a Comment
Read more

Class 12 Maths Chapter 2 Inverse Trigonometric Functions

Inverse Trigonometric Functions 
Post a Comment
Read more