limit integration(1)/1+(x+2)(x+1)^(1/2))
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Answered by
1
Answer:
We know that
∫
a
b
f(x)dx=(b−a)
n→∞
lim
n
1
(f(a)+f(a+h)+...+f(a+(n−1)h))
Putting a=0,b=2,h=
n
b−a
=
n
2−0
=
n
2
in ∫
0
2
x
2
+1dx
I=(2−0)
n→∞
lim
n
1
(f(0)+f(n)+f(2n)+...+f((n−1)h))
f(0)=1
f(h)=h
2
+1
=(
n
2
4
)+1
f((n−1)h)=(n−1)
2
×
n
2
4
+1
∴I=2
n→∞
lim
n
1
((1+1+...ntimes)+(0+
n
2
4
+
n
2
16
+...+
n
2
(n−1)
2
))
=2
n→∞
lim
n
1
(n+
n
4
6
(n−1)n(2n−1)
)
=2
n→∞
lim
(1+
3
2
(1−
n
1
)(2−
n
1
))
=2×(1+
3
4
)=
3
14
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Answer:
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