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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

NCERT Solutions for integrals Class 12 exercises 7.11 are accessible in PDF format which is available at CoolGyan’s online learning portal. The solutions to this particular exercise from chapter 7 Integrals have been prepared by our experts who have years of experience in this field. Students can find, well-explained solutions to maths problems at CoolGyan’s online learning portal. All the solutions are provided in a way that will help you in understanding the concept effectively. You can download them in PDF format for free.

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 = 

=[(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)

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 = 

π2log21π2log2−1

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+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)

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.