## 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 = I_{1} – ……….(v)

Where I_{1} = …….(vi)

Putting in eq. (vi),

Limits of integration when and

From eq. (vi),

I_{1} =

=

=

I_{1} =

=

I_{1} = [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 x^{3} , xcosx and tan^{5}x 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.