TargetIIT

[im]http://latex.codecogs.com/gif.latex?Lim\;%20n\to%20\infty%20\;%20\frac{1}{n^{2}}%20\;%20\sum_{k=0}^{n-1}\:%20\:%20\begin{bmatrix}%20k\int_{k}^{k+1}%28%28x-k%29%28k+1-x%29%29^{1/2}%20\end{bmatrix}[/im]

A) π/32
B) π/16
C) π/8
D) π/4

We have
[im]http://codecogs.izyba.com/gif.latex?%28x-k%29%28k+1-x%29%3D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E2-%5Cleft%28x-%5Cdfrac%7B2k+1%7D%7B2%7D%5Cright%29%5E2[/im]
Hence,
[im]http://codecogs.izyba.com/gif.latex?%5Cint_k%5E%7Bk+1%7D%5Csqrt%7B%28x-k%29%28k+1-x%29%7D%5C%20%5Cmathrm%20dx%3D%5Cint_k%5E%7Bk+1%7D%5Csqrt%7B%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E2-%5Cleft%28x-%5Cdfrac%7B2k+1%7D%7B2%7D%5Cright%29%5E2%7D%5C%20%5Cmathrm%20dx[/im]
Apply the substitution
[im]http://codecogs.izyba.com/gif.latex?x-%5Cdfrac%7B2k+1%7D%7B2%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Csin%5Ctheta%5Cquad%5CRightarrow%5C%20%5Cmathrm%20dx%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccos%5Ctheta%20%5C%20%5Cmathrm%20d%5Ctheta[/im]
Also when x=k, sinθ = -1, i.e. θ=-π/2 and when x=k+1, sinθ=1, i.e. θ=π/2. Hence the given integral is [im]http://codecogs.izyba.com/gif.latex?%5Cint_%7B-%5Cpi/2%7D%5E%7B%5Cpi/2%7D%5Cdfrac%7B1%7D%7B4%7D%5Ccos%5E2%5Ctheta%20%5C%20%5Cmathrm%20d%5Ctheta%20%3D%5Cdfrac%7B%5Cpi%7D%7B8%7D[/im]
Hence, the required limit becomes
[im]http://codecogs.izyba.com/gif.latex?%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cdfrac%7B1%7D%7Bn%5E2%7D%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D%5Cdfrac%7Bk%5Cpi%7D%7B8%7D%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cdfrac%7B1%7D%7Bn%5E2%7D%20%5Cdfrac%7B%5Cpi%7D%7B8%7D%5Cdfrac%7Bn%28n-1%29%7D%7B2%7D%3D%5Cdfrac%7B%5Cpi%7D%7B16%7D[/im]