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:-Interjection Exercises With Answers Sheet-1

Interjections An interjection is a word that expresses sudden feelings: Joy, grief, surprise, praise, anger etc. Examples: 1. Hello! How are you? 2. Hurrah! We have won the match. 3. Alas! My granny is dead! 4. Hush! The baby is asleep. 5. Hurrah! We have won the tennis match. 6. Oh! I forgot to tell you something. 7. Ah! What a pleasant surprise. 8. Oh God! I got such fright. 9. Bravo! Well done. Interjections and their Expressions: Interjection Expression Hurrah! Joy Oh dear! borrow Alas! Ah! Sorrow Ugh! disgust Oh! Wonder Good heavens! horror Interjection Expression Bravo! Praise for goodness sake! anger Fie! Hatred Hello! address My goodness! Surprise Wow! Happiness Match the following columns: Column-A Column-B Hurrah! Sorrow Oh! Happiness Hello! Wonder Wow! Joy Alas! Address Fill in the blanks: 1. _____! How can you behave like this? 2. ______! I have secured first position in the class. 3. ______! I have won a prize. 4. ______! Did you hear me? 5. _______! ...
Post a Comment
Read more

GRAMMAR:-CONJUNCTION EXERCISES WITH ANSWERS SHEET-2

1. ______, you've got a chance, you might as well make full use of it. (A) Now that (B) After (C) Although (D) As soon as 2. She has not spoken to us _____ we had the argument. (A) as (B) so (C) while (D) since 3. _____ you refuse to pay the ransom, the kidnappers might hurt the child. (A) Lest (B) Unless (C) If (D) Or else 4. _____ he was about to fall asleep the telephone rang. (A) As (B) Since (C) Unless (D) So that 5. _____ the couple goes, their children follow them. (A) When (B) Where (C) If (D) Wherever 6. Father locked the garage door _____ no one could tamper with the car. (A) because (B) so that (C) for (D) lest 7. It was only when I reread his po...
Post a Comment
Read more

Force

What is Force? Force can be defined as a push or a pull that changes or tends to change the state of rest or uniform motion of an object or changes the direction or shape of an object.  It causes objects to accelerate or add to their overall pressure. In simple terms, it is a push or a pull on an object that takes place when two objects interact. It is the basic cause of motion from rest. Unit of Measurement It is measured in ‘Newton’ (N). Newton measures the force needed to move, accelerate or speed up objects. The basic formula for it is, F = m.a, where, ‘m’ stands for the mass in kilograms and ‘a’ stands for acceleration in m/sec 2 .  It may also be measured in Pounds. Force is a Vector Quantity Types of Forces There are different types of forces in the Universe. Based on the nature of the interaction between two bodies, they  can be classified into two main categories viz. Contact Forces. Non-Contact Forces (Field Forces or Action at a D...
Post a Comment
Read more