# NCERT Solutions class 12 Maths Exercise 7.11 (Ex 7.11) Chapter 7 Integrals

## NCERT Solutions for Class 12 Maths Exercise 7.11 hapter 7 Integrals – FREE PDF Download

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# NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.11) Exercise 7.11

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 6.

1.

Ans. Let I =  ……….(i)

……….(ii)

2I =

=>2I=(x)π20=>2I=(x)0π2

2I =

I =  Ans.

2.

Ans. Let I =   ………..(i)

I =

I =  ……….(ii)

2I =

2I =

=>2I=(x)π20=>2I=(x)0π2

2I =

I =  Ans.

3.

Ans. Let I =   ………..(i)

I =

I =   ……….(ii)

2I =

2I =

=>2I=(x)π20=>2I=(x)0π2

2I =

I =  Ans.

4.

Ans. Let I =  ………..(i)

I =

I =  ……….(ii)

2I =

2I =

=>2I=(x)π20=>2I=(x)0π2

2I =

I =  Ans.

5.

Ans. Let I =  ……….(i)

Putting

From eq. (i),

I =

=[(424)(25210)]+[(252+10)(424)]=−[(42−4)−(252−10)]+[(252+10)−(42−4)]

Ans.

6.

Ans. Let I =   ……….(i)

Putting

From eq. (i),

I =

=(x225x)52+(x225x)85=−(x22−5x)25+(x22−5x)58

=[(25225)(210)]+[(3240)(25225)]=−[(252−25)−(2−10)]+[(32−40)−(252−25)]

=(252+8)+(8+252)=−(−252+8)+(−8+252)

=25288+252=252−8−8+252

= 25 – 16

= 9 Ans.

By using the properties of definite integrals, evaluate the integrals in Exercises 7 to 11.

7.

Ans. Let I =

I =

I =

Ans.

8.

Ans. Let I =         ……….(i)

I =

I =

…..(ii)

2I =

2I =

I =  Ans

9.

Ans. Let I =

I =

I =

Ans.

10.

Ans. Let I =

……….(i)

I =

I =         ……….(ii)

2I =

2I =

I =

π2log21π2log2−1

π2log12π2log12 Ans.

11.

Ans. Let I =

…(i)

I =

I =         ……….(ii)

2I =

I =  Ans.

Using properties of definite integrals, evaluate the following integrals in Exercise 12 to 18.

12.

Ans. Let I =          ……….(i)

I =

……….(ii)

2I =

2I =

2I =

2I =

I =

Ans.

13.

Ans. Let I =

Here

is an odd function of

I =  = 0

14.

Ans.

Here

15.

Ans. Let I =  ……….(i)

I =

….(ii)

Adding eq. (i) and (ii), we have 2I = 0   I = 0 Ans.

16.

Ans. Let I =         ……….(i)

I =

……….(ii)

2I =

2I =

I =  ……….(iii)

I =

……….(iv)

2I =

I =

I =

I = I1 –  ……….(v)

Where I1 =  …….(vi)

Putting  in eq. (vi),

Limits of integration when  and

From eq. (vi),

I1 =

I1 =

I1 =  [From eq. (iii)]

Putting this value in eq. (v), I =

2I =

2I – I =

I =

17.

Ans. Let I =  ……….(i)

I =

……….(ii)

2I =

=  =

I =   Ans.

18.

Ans. Let I =          ……….(i)

Here

From eq. (i),

I =

I =

=12+84+12=12+8−4+12

= 9 – 4

= 5 Ans.

19. Show that 0af(x)g(x)dx=20af(x)dx∫0af(x)g(x)dx=2∫0af(x)dx if f and g are defined as  and

Ans. Here  ……..(i)  and   ………(ii)

Let I =  ……….(iii)

I =

[From eq. (i)] ……….(iv)

2I =

2I =  [From eq. (ii)]

2I =

I =

Hence proved.

Choose the correct answer in Exercises 20 and 21.

20. The value of  is:

(A) 0

(B) 2

(C)

(D) 1

Ans. Let I =

I =

Because x3 , xcosx and tan5x all are odd functions.

Therefore, option (C) is correct.

21. The value of  is:

(A) 2

(B)

(C) 0

(D)

Ans. Let I =  ……….(i)

I =
……….(ii)
2I =

2I =
= 0
I = 0
Option (C) is correct.