Class 12 Maths NCERT Solutions for Chapter 7 Integrals Exercise 7.9
![Class 12 Maths NCERT Solutions for Chapter 7 Integrals Exercise 7.9 Class 12 Maths NCERT Solutions for Chapter 7 Integrals Exercise 7.9](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhWGtE72oCYbHGNg3LwQ-2BtorpeoK-u8lI9dzvu-ki9VoEokx036mC-iCzQc2Qhap-zT5m-cQCVZ6jMxQgG-xj8BPlE433ciDDGlXpTxOTQPkC8hMlpQza3rxvv2_Bi4qm24txGAhi9jL9a76br-JrgCKuyt_8Db1_H851QvPfOWN_xPUyQBH2DYUL/w657-h311-rw/cbse-class-12-maths-ncert-solutions-chapter-7-exercise-7-9.jpg)
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)
= 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)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiFXdsPwDvf4Vz9K6zhgf74GpHTcVC2LHJy7nPWAEu8rTlkLNgLAwT5SlE6DySBs9-yZI2wa_4GpkJ7zQ_W_HVtV0ObvTdkNHzKpYeJDkZK1NDfoC-RsYKyRkLnSkdmdcj-CkPs_QRtMnBv09kdEYzaI10OKVyw3EvH_UM8XLdB_BEZEMjokmG3LGmE/w438-h159-rw/NCERT%20Solutions%20for%20Chapter%207%20Integrals%20Class%2012%20Maths%20Exercise%20-%207.9%20img%203.JPG)
= 33 - (35/3)
= (99 - 35)/3
= 64/3
= 33 - (35/3)
= (99 - 35)/3
= 64/3
4. Evaluate the definite integrals ∫0x/4 sin2x dx
Solution
Let I = ∫0x/4 sin2x dx
∫sin 2x dx = (- cos 2x/2) = F(x)
By second fundamental theorem of calculus, we obtain
∫sin 2x dx = (- cos 2x/2) = F(x)
By second fundamental theorem of calculus, we obtain
5. Evaluate the definite integrals ∫0x/2 cos 2x dx
Solution
Let I = ∫0x/2 cos 2x dx
∫cos 2x dx = (sin 2x/2) = F(x)
By second fundamental theorem of calculus, we obtain
∫cos 2x dx = (sin 2x/2) = F(x)
By second fundamental theorem of calculus, we obtain
6. Evaluate the definite integrals ∫45 ex dx
Solution
Let I = ∫45 ex dx
∫ex dx = ex = f(x)
By second fundamental theorem of calculus, we obtain
By second fundamental theorem of calculus, we obtain
I = F(5) - F(4)
= e5 - e4
= e5 - e4
= e4 (e - 1)
7. Evaluate the definite integrals ∫0x/4 tan x dx
Solution
Let I = ∫0Ï€/4 tan x dx
∫tan x dx = -log |cos x| = F(x)
By second fundamental theorem of calculus, we obtain
By second fundamental theorem of calculus, we obtain
8. Evaluate the definite integrals ∫Ï€/6 Ï€/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)
= (Ï€/2) - 0
= π/2
= sin-1 (1) - sin-1 (0)
= (Ï€/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
= tan-1 (1) - tan-1 (0)
= π/4
11. Evaluate the definite integrals ∫23 dx/(x2 - 1)
Solution
12. Evaluate the definite integrals ∫0Ï€/4 cos2x 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 xex2 dx
Solution
Let I = ∫01 xex2 dx
Put x2 = t
Put x2 = t
⇒ 2x dx = dt
As x → 0, t → 0 and as x → 1, t →1,
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjIskH_yJHWsbm0wCE390dbCbBDLHnIYp99gXDozDFs2l_8P_oCVLsD9G-svBycRhs3vREaMS83LBqjg_VDuGWx5Alju6vEM7fOM38SyP8ikoVwvplZX9CwS0KmsjNIo7NSfoSYwb82MsI7b5PJxEK90aItusbOtCSSZvKMn_QocpnsSFeiQ88MdQK2/w149-h89-rw/NCERT%20Solutions%20for%20Chapter%207%20Integrals%20Class%2012%20Maths%20Exercise%20-%207.9%20img%2015.JPG)
By second fundamental theorem of calculus, we obtain
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)
= (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 ∫0Ï€/4 (2 sec2 x + x3 + 2)dx
Solution
18. Evaluate the definite integrals ∫0Ï€ (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(Ï€) - F(0)
= sinπ - sin 0
= 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)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg6NvTYCCyPYYRWi0OFO_1V9m_uODUXZuqQdbZZtnYc64ea1H55pgzvQWBUzXdUMK3GC1qJ5QURaTEQvoL1b8wiYl70U0JTEiueAJCb1jafHi_jH0ep3F0l6PUvZfczGOVUKab5WlK5-iyrOOeTxNzF6HgDFlEs7qCiHdyH7QsnoaqS53lK_bEHzOOl/w452-h257-rw/NCERT%20Solutions%20for%20Chapter%207%20Integrals%20Class%2012%20Maths%20Exercise%20-%207.9%20img%2022.JPG)
20. Evaluate the definite integrals ∫01 [xex + sin(Ï€x/4)]
Solution
21. Choose the correct answer ∫1√3 dx/(1 + x2 ) equals
(A) π/3
(B) 2Ï€/3
(C) π/6
(D) π/12
(B) 2Ï€/3
(C) π/6
(D) π/12
Solution
∫dx/(1 + x2 ) = tan-1 x = F(x)
By second fundamental theorem of calculus, we obtain
By second fundamental theorem of calculus, we obtain
= tan-1 √3 - tan-1 1
= π/3 - π/4
= π/12
= π/12
Hence, the correct answer is D.
22. Choose the correct answer ∫02/3 dx/(4 + 9x2 ) equals is
Solution
By second fundamental theorem of calculus, we obtain
Hence, the correct answer is C.