NCERT Solutions for Chapter 7 Integrals Class 12 Maths Exercise 7.9

Class 12 Maths NCERT Solutions for Chapter 7 Integrals Exercise 7.9

Integrals Exercise 7.9 Solutions

1. Evaluate the definite integrals ∫_-1¹_­ (x + 1) dx 

Solution

Let I =  ∫_-1¹_­ (x + 1) dx 
∫(x + 1)dx = x²_­/2 + x = F(x) 
By second fundamental theorem of calculus, we obtain 
I = F(1) - F(-1)
= (1/2 + 1) - (1/2 - 1)
= (1/2 + 1 - 1/2 + 1)
= 2

2. Evaluate the definite integrals  ∫₂³_­ (1/x) dx 

Solution

∫(1/x) dx = log |x| = F(x) 
By second fundamental theorem of calculus, we obtain 

I  = F(3) - F(2) 
= log|3| - log|2| = log (3/2) 

3. Evaluate the integrals in using substitution ∫₀¹_­ sin^-1_­ [2x/(1 + x²_­ ) dx 

Solution

By second fundamental theorem of calculus, we obtain 

I = F(2) - F(1) 

= 33 - (35/3)
= (99 - 35)/3 
= 64/3 

4. Evaluate the definite integrals ∫₀^x/4_­ sin2x dx 

Solution

Let I = ∫₀^x/4_­ sin2x dx 
∫sin 2x dx = (- cos 2x/2) = F(x) 
By second fundamental theorem of calculus, we obtain

5. Evaluate the definite integrals ∫₀^x/2_­ cos 2x dx 

Solution

Let I =  ∫₀^x/2_­ cos 2x dx 
∫cos 2x dx = (sin 2x/2) = F(x) 
By second fundamental theorem of calculus, we obtain 

6. Evaluate the definite integrals ∫₄⁵_­ e^x_­ dx 

Solution

Let I =  ∫₄⁵_­ e^x_­ dx

∫e^x_­ dx = e^x_­ = f(x) 
By second fundamental theorem of calculus, we obtain 

I = F(5) - F(4)
= e⁵_­ - e⁴_­ 

= e⁴_­ (e - 1) 

7. Evaluate the definite integrals ∫₀^x/4_­ tan x dx   

Solution

Let I = _­ ∫₀^π/4_­ tan x dx

∫tan x dx = -log |cos x| = F(x) 
By second fundamental theorem of calculus, we obtain 

8. Evaluate the definite integrals ∫_π/6 ^π/4 cosec x dx   

Solution

∫(cosec x) dx = log | cosec x - cot x| = F(x) 
By second fundamental theorem of calculus, we obtain 

9. Evaluate the definite integrals ∫₀⁴ dx/(√ 1 - x²_­ ) 

Solution

By second fundamental theorem of calculus, we obtain 

I = F(I) - F(0) 
= sin^-1_­ (1) -  sin^-1_­ (0) 
= (π/2) - 0 
= π/2 

10. Evaluate the definite integrals ∫₀⁴  dx/(1 + x²_­ )

Solution

By second fundamental theorem of calculus, we obtain 

I = F(1) - F(0)
= tan^-1_­ (1) - tan^-1_­ (0) 
= π/4 

11. Evaluate the definite integrals ∫₂³ dx/(x²_­ - 1)

Solution

12. Evaluate the definite integrals ∫₀^π/4_­ cos²_­x dx 

Solution

13. Evaluate the definite integrals ∫₂³ xdx/(x²_­ + 1) 

Solution

14. Evaluate the definite integrals ∫₀¹ (2x + 3)/(5x²_­ + 1) dx 

Solution

15. Evaluate the definite integrals  ∫₀¹ xe^x2 dx 

Solution

Let I = ∫₀¹ xe^x2 dx  
Put x²_­ = t

⇒ 2x dx = dt 
As x → 0, t → 0 and as x → 1, t →1, 

By second fundamental theorem of calculus, we obtain 

I = F(1) - F(0) 
= (1/2)e - (1/2)e⁰_­
= (1/2) (e - 1)

16. Evaluate the definite integrals  ∫₁² 5x²_­ /(x²_­ + 4x + 3) 

Solution 

Equating the coefficients of x and constant term, we obtain
A = 10 and B = -25 

Substituting the value of I₁ in (1) , we obtain

17. Evaluate the definite integrals ∫₀^π/4_­ (2 sec²_­ x + x³_­ + 2)dx 

Solution

18. Evaluate the definite integrals ∫₀^π_­ (sin²_­ x/2 - cos²_­ x/2) dx 

Solution

∫(cos x) dx = sin x = F(x) 
By second fundamental theorem of calculus, we obtain 

I = F(π) - F(0) 
= sinπ - sin 0 
= 0 

19. Evaluate the definite integrals  ∫₀² [(6x + 3)/(x²_­ + 4)] dx 

Solution

By second fundamental theorem of calculus, we obtain 

I = F(2) - F(0) 

20. Evaluate the definite integrals  ∫₀¹ [xe^x_­ + sin(πx/4)] 

Solution

21. Choose the correct answer  ∫₁^√3 dx/(1 + x²_­ ) equals 

(A) π/3
(B) 2π/3
(C) π/6 
(D) π/12

Solution

∫dx/(1 + x²_­ ) = tan^-1_­ x = F(x) 
By second fundamental theorem of calculus, we obtain 

= tan^-1_­ √3 - tan^-1_­ 1 

= π/3 - π/4 
= π/12 

Hence, the correct answer is D. 

22. Choose the correct answer ∫₀^2/3_­ dx/(4 + 9x²_­ ) equals is 

Solution

By second fundamental theorem of calculus, we obtain 

Hence, the correct answer is C.