Question

______________________??​

Answer 1

$\lim _{n\to \infty }\left(\frac{1}{n^2}\right)\sum _{k=0}^{n-1}\:\left[k\int _k^{k+1}\:\sqrt{\left(x-k\right)\left(k+1-x\right)}dx\right]$

$Steps$*

$\lim _{n\to \infty \:}\left(\frac{1}{n^2}\cdot \sum _{k=0}^{n-1}k\cdot \int _k^{k+1}\sqrt{\left(x-k\right)\left(k+1-x\right)}dx\right)$

$\int _k^{k+1}\sqrt{\left(x-k\right)\left(k+1-x\right)}dx=\frac{\pi }{8}$

$=\lim _{n\to \infty \:}\left(\frac{1}{n^2}\cdot \sum _{k=0}^{n-1}k\frac{\pi }{8}\right)$

$\sum _{k=0}^{n-1}k\frac{\pi }{8}=\frac{\pi n\left(n-1\right)}{16}$

$=\lim _{n\to \infty \:}\left(\frac{1}{n^2}\cdot \frac{\pi n\left(n-1\right)}{16}\right)$

$\lim _{n\to \infty \:}\left(\frac{1}{n^2}\cdot \frac{\pi n\left(n-1\right)}{16}\right)=\frac{\pi }{16}$

$=\frac{\pi }{16}$

Answer 2

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