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)
Adding eq. (i) and (ii),
2I = 
= 
=>2I=(x)π20=>2I=(x)0π2
2I = 
I = 
Ans.
2. 
Ans. Let I = 
………..(i)
I = 
I = 
……….(ii)
Adding eq. (i) and (ii),
2I = 
= 
2I =
=>2I=(x)π20=>2I=(x)0π2
2I =
I = Ans.
3. 
Ans. Let I = 
………..(i)
I = 
I = 
……….(ii)
Adding eq. (i) and (ii),
2I = 
= 
2I =
=>2I=(x)π20=>2I=(x)0π2
2I =
I = Ans.
4. 
Ans. Let I = 
………..(i)
I = 
I = 
……….(ii)
Adding eq. (i) and (ii),
2I = 
= 
2I =
=>2I=(x)π20=>2I=(x)0π2
2I =
I = Ans.
5. 
Ans. Let I = ……….(i)
Putting 
From eq. (i),
I = 
= 
= 
=−[(42−4)−(252−10)]+[(252+10)−(42−4)]=−[(42−4)−(252−10)]+[(252+10)−(42−4)]
= 
= 
= 
Ans.
6. 
Ans. Let I = ……….(i)
Putting 
From eq. (i),
I = 
= 
=−(x22−5x)52+(x22−5x)85=−(x22−5x)25+(x22−5x)58
=−[(252−25)−(2−10)]+[(32−40)−(252−25)]=−[(252−25)−(2−10)]+[(32−40)−(252−25)]
=−(−252+8)+(−8+252)=−(−252+8)+(−8+252)
=252−8−8+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)
Adding eq. (i) and (ii),
2I = 
= 
2I = 
= 
= 
I = 
Ans
9. 
Ans. Let I =
= 
I = 
= 
= 
= 
I = 
= 
= 
Ans.
10. 
Ans. Let I =
= 
= 
= 
= 
……….(i)
I = 
I = 
……….(ii)
Adding eq. (i) and (ii),
2I = 
= 
2I = 
= 
= 
= 
I = 
= 
= 
= π2log2−1π2log2−1
= π2log12π2log12 Ans.
11. 
Ans. Let I =
= 
…(i)
I = 
I = 
……….(ii)
Adding eq. (i) and (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)
Adding eq. (i) and (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)
Adding eq. (i) and (ii),
2I = 
= 
2I = 
= 
= 
I = 
……….(iii)
I = 
= 
……….(iv)
Adding eq. (i) and (ii),
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)
Adding eq. (i) and (ii),
2I = 
= 
= 
= 
= 
I = 
Ans.
18. 
Ans. Let I = ……….(i)
Here 
From eq. (i),
I = 
= 
I = 
= 
=12+8−4+12=12+8−4+12
= 9 – 4
= 5 Ans.
19. Show that ∫0af(x)g(x)dx=2∫0af(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)
Adding eq. (iii) and (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)
Adding eq. (i) and (ii),
2I = 
= 
2I = 
= 
= 0
I = 0
Option (C) is correct.