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CBSE Class 12 Math 2018 Solved Paper

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Question : 27
Total: 29
Evaluate :
Ï€
4
∫
0
sinx+cosx
16+9sin2x
 dx
OR
Evaluate :
3
∫
1
(x2+3x+ex) dx
as the limit of the sum
Solution:  
Ï€
4
∫
0
sinx+cosx
16+9sin2x
 dx
Now, sin2x = 2sinx cosx
∴ 1- sin2x = 1 - 2sinx cosx
∴ 1 - sin2x = (sinx−cosx)2
Put sin x - cos x = t
⇒ (sin x + cos x) dx = dt
x =
Ï€
4
 , t = 0
x = 0 , t = - 1
So,
0
∫
−1
1
16−9(1−t2)
 dt
=
0
∫
−1
1
16−9−9t2
 dt
=
0
∫
−1
1
25−9t2
 dt
=
1
9
0
∫
−1
1
25
9
−t2
 dt
=
(
1
9
1
2×
5
3
‌log|
5
3
+t
5
3
−t
|)−10

=
1
30
 (log|1|−log‌
1
4
)
=
1
30
 log 4
OR
Given
3
∫
1
(x2+3x+ex) dx
⇒ a = 1 , b = 3
⇒ h =
3−1
n
 and f (x) = x2 + 3x + ex
I =
3
∫
1
(x2+3x+ex) dx
I =
lim
h→0
 h [f (1) + f (1 + h) + ... + f (1 + (n - 1)h)]
I =
lim
h→0
 h [4 + e + [(1+h)2 + 3 (1 + h) + e1+h]] + [(1+2h)2 + 3 (1 + 2h) + e1+2h] + ... + [(1+(n−1)h)2 + 3 (1 + (n - 1) h + e1+(n−1)h)]
I =
lim
h→0
 h
[4n+e+h2
(n−1)(n−2)(n−3)
6
+2h
n(n−1)
2
+3h
n(n−1)
2
+3h
n(n−1)
2
+e
[eh(eh(n−1)−1)]
eh−1
]

I =
lim
n→∞
 4n ×
2
n
 + e ×
2
n
+
8
n3
(n−1)(n−2)(n−3)
6
 +
5×4
n2
(
n2−n
2
) + e
2
n+1
 (e
2
n
(n−1)−1)
I = 8 +
8
3
 + 10 + e (e2 - 1)
I =
62
3
 + e3 - e
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