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CBSE Class 12 Math 2012 Solved Paper

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Question : 26
Total: 29
Prove that
π
4
∫
0
√tanx + √cotx dx = √2,
π
2
OR
Evaluate
3
∫
1
2x2 + 5x dx as a limit of sum.
Solution:  
π
4
∫
0
√tanx + √cotx dx
=
π
4
∫
0
 (
√sinx
√cosx
+
√cosx
√sinx
) dx
=
π
4
∫
0
 (
sinx+cosx
√sinxcosx
) dx
= √2
π
4
∫
0
 (
sinx+cosx
√2sinxcosx
) dx
√2
π
4
∫
0
 (
sinx+cosx
√1−sinx−cosx2
) dx
Put sin x - cos x = t ⇒ (cos x + sin x) dx = dt
If x = 0 , t = 0 - 1 = - 1
and if x =
π
4
 , t =
1
√2
−
1
√2
 = 0
∴
π
4
∫
0
 √tanx+√cotx dx = √2‌
0
∫
−1
dt
√1−t2
= √2|sin−1t|−10
= √2|sin−10−sin−1(−1)|
= √2|0+
π
2
|
= √2×
π
2
OR
3
∫
1
2x2 + 5x dx
Here, a = 1 , b = 3 , f (x) = 2x2 + 5x
∴ nh = b - a = 3 - 1 = 2
Now
b
∫
a
 f (x) dx =
lim
h→0
 f (a) + f (a + h) + f (a + 2h) + ... + f (a + (n - 1)h)
∴
3
∫
1
2x2 + 5x dx
=
lim
h→0
 h |2 (1)2 +5 (1) + 2 (1+h)2 + 5 (1 + h) + [2(1+2h)2 + 5 (1 + 2h)] ... + [2 (1+(n−1)h)2 + 5 (1 + (n - 1) h)]|
=
lim
h→0
 |7 + (2h2 + 9h + 7) + (8h2 + 18h + 7) + ... + (2 (n−1)2h2 + 9 (n - 1) h + 7)|
=
lim
h→0
 |7n + 2h2 (12+2+...+(n−1)2) + 9h (1 + 2 + ... + (n - 1))|
= li
m
h→0
|7n+2h2
n(n−1)(2n−1)
6
+9h
n(n−1)
2
|

=
lim
h→0
|7nh+2
nh(nh−2)(2nh−h)
6
+9
nh(nh−h)
2
|

=
lim
h→0
|14+2
2(2−h)(4−h)
6
+9
2(2−h)
2
|

= 14 +
16
3
 + 18 =
112
3
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