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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


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