NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10.
- Integrals Class 12 Ex 7.1
- Integrals Class 12 Ex 7.2
- Integrals Class 12 Ex 7.3
- Integrals Class 12 Ex 7.4
- Integrals Class 12 Ex 7.5
- Integrals Class 12 Ex 7.6
- Integrals Class 12 Ex 7.7
- Integrals Class 12 Ex 7.8
- Integrals Class 12 Ex 7.9
- Integrals Class 12 Ex 7.11
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 12 |
| Subject | Maths |
| Chapter | Chapter 7 |
| Chapter Name | Integrals |
| Exercise | Ex 7.10 |
| Number of Questions Solved | 10 |
| Category | NCERT Solutions |
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10
Evaluate the integrals in Exercises 1 to 8 using substitution.
Ex 7.10 Class 12 Maths 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
\(\therefore I=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { dt }{ t } } ={ \left[ \frac { 1 }{ 2logt } \right] }_{ 1 }^{ 2 }\quad =\frac { 1 }{ 2 } log2\)
Ex 7.10 Class 12 Maths 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
Ex 7.10 Class 12 Maths 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 => \(\theta \frac { \pi }{ 4 } \)
\(\frac { 1 }{ 2 }\)
Ex 7.10 Class 12 Maths 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
Ex 7.10 Class 12 Maths 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
\(\therefore I=\int _{ 1 }^{ 0 }{ \frac { -dt }{ 1+{ t }^{ 2 } } =-{ \left[ { tan }^{ -1 }t \right] }_{ 1 }^{ 0 } } =\frac { \pi }{ 4 } \)
Ex 7.10 Class 12 Maths Question 6.
\(\int _{ 0 }^{ 2 }{ \frac { dx }{ x+4-{ x }^{ 2 } } =I } \)
Solution:
\(\int _{ 0 }^{ 2 }{ \frac { dx }{ x+4-{ x }^{ 2 } } =I } \)
Ex 7.10 Class 12 Maths Question 7.
\(\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+5 } =I } \)
Solution:
\(I=\int _{ -1 }^{ 1 }{ \frac { dx }{ { (x+1) }^{ 2 }+{ 2 }^{ 2 } } } =\frac { 1 }{ 2 } { \left[ { tan }^{ -1 }\frac { x+1 }{ 2 } \right] }_{ -1 }^{ 1 }\quad =\frac { \pi }{ 8 } \)
Ex 7.10 Class 12 Maths 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
Ex 7.10 Class 12 Maths 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 = \(\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { { { (x-x }^{ 3 }) }^{ \frac { 1 }{ 3 } } }{ { x }^{ 4 } } dx } \quad =\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { { x }^{ \frac { 1 }{ 3 } }(1-{ x }^{ 2 })^{ \frac { 1 }{ 3 } } }{ { x }^{ 4 } } dx } \)
Ex 7.10 Class 12 Maths 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
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