_____________________!!
Answers
Answer:
\lim _{n\to \infty }(\frac{1}{n^2})\sum _{k=0}^{n-1}\:[k\int _k^{k+1}\:\sqrt{(x-k)(k+1-x)}dx]lim
n→∞
(
n
2
1
)∑
k=0
n−1
[k∫
k
k+1
(x−k)(k+1−x)
dx]
StepsSteps *
\lim _{n\to \infty \:}(\frac{1}{n^2}\cdot \sum _{k=0}^{n-1}k\cdot \int _k^{k+1}\sqrt{(x-k)(k+1-x)}dx)lim
n→∞
(
n
2
1
⋅∑
k=0
n−1
k⋅∫
k
k+1
(x−k)(k+1−x)
dx)
\int _k^{k+1}\sqrt{(x-k)(k+1-x)}dx=\frac{\pi }{8}∫
k
k+1
(x−k)(k+1−x)
dx=
8
π
=\lim _{n\to \infty \:}(\frac{1}{n^2}\cdot \sum _{k=0}^{n-1}k\frac{\pi }{8})=lim
n→∞
(
n
2
1
⋅∑
k=0
n−1
k
8
π
)
\sum _{k=0}^{n-1}k\frac{\pi }{8}=\frac{\pi n(n-1)}{16}∑
k=0
n−1
k
8
π
=
16
πn(n−1)
=\lim _{n\to \infty \:}(\frac{1}{n^2}\cdot \frac{\pi n(n-1)}{16})=lim
n→∞
(
n
2
1
⋅
16
πn(n−1)
)
\lim _{n\to \infty \:}(\frac{1}{n^2}\cdot \frac{\pi n(n-1)}{16})=\frac{\pi }{16}lim
n→∞
(
n
2
1
⋅
16
πn(n−1)
)=
16
π
=\frac{\pi }{16}=
16
π
refee you the attachment
