Test Index

CBSE Class 12 Maths 2010 Solved Paper

© examsnet.com
Question : 23
Total: 29
Evaluate
3
∫
1
(3x2+2x) dx as limit of sums.
OR
{(x,y):
x2
9
+
y2
4
≤1≤
x
3
+
y
2
}
Solution:  
I =
3
∫
1
(3x2+2x) dx
Here a = 1 , b = 3
f (x) = 3x2 + 2x
h =
b−a
n
 =
2
n
Since,
b
∫
a
 f (x) dx =
lim
h→0
 h [f (a) + f (a + h) + ... + f (a + (n - 1) h)]
So,
3
∫
1
(3x2+2x) =
lim
h→0
 h [3 (1)2 + 2 (1)) + (3 (1+h)2 + 2 (1 + h) + 3 (1+2h)2 + 2 (1 + 2h)) ... + 3 (1+(n−1)h)2 + 2 (1 + (n - 1) h)]
=
lim
h→0
 h [3 (n) + 3 (h2+4h2+..(n−1)2h2) + 3 (2h + 4h + ... + 2 (n - 1) h) + 2n + 2 (h + 2h + ... + (n - 1) h)]
=
lim
h→0
 [5nh + 3h3(12+22+...+(n−1)2) + 6h2 (1 + 2 + ... + (n - 1)) + 2h2 (1 + 2 + (n - 1))]
=
lim
h→0
[5nh+3h2×
(n−1)n(2n−1)
6
+
8h2(n)(n−1)
2
]

=
lim
h→0
[10+
nh−h)‌n‌h(2nh−h)
2
+4(nh)(nh−h)]

= [10+
2×2×4
2
+4×2×2]
= 10 + 8 + 16 = 34
OR
⇒
x2
9
+
y2
4
 = 1
⇒ y =
2
3
√9−x2
Given line
x
3
+
y
2
 = 1
⇒ y = (2−
2x
3
)
Required Area {(x,y):
x2
9
+
y2
4
≤1≤
x
3
+
y
2
} is given below

Required Area =
3
∫
0
(y1−y2) dx
=
3
∫
0
[
2
3
√9−x2−(2−
2x
3
)] dx
=
[
2
3
(
x
2
√9−x2+
3
2
sin−1
x
3
)−2x+
x2
3
]03

= [
2
3
(
9
2
sin−11)−6+3] - 0
= 3 ×
Ï€
2
 - 3 =
3
2
 (Ï€ - 2) sq units
© examsnet.com
Go to Question:
Prev Question
Next Question