Given
we are given a function
[tex]f(x)=x^2+5[/tex]over the interval [0,5].
Required
we need to find formula for Riemann sum and calculate area under the curve over [0,5].
Explanation
If we divide interval [a,b] into n equal intervals, then each subinterval has width
[tex]\Delta x=\frac{b-a}{n}[/tex]and the endpoints are given by
[tex]a+k.\Delta x,\text{ for }0\leq k\leq n[/tex]For k=0 and k=n, we get
[tex]\begin{gathered} x_0=a+0(\frac{b-a}{n})=a \\ x_n=a+n(\frac{b-a}{n})=b \end{gathered}[/tex]Each rectangle has width and height as
[tex]\Delta x\text{ and }f(x_k)\text{ respectively.}[/tex]we sum the areas of all rectangles then take the limit n tends to infinity to get area under the curve:
[tex]Area=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)[/tex]Here
[tex]f(x)=x^2+5\text{ over the interval \lbrack0,5\rbrack}[/tex][tex]\Delta x=\frac{5-0}{n}=\frac{5}{n}[/tex][tex]x_k=0+k.\Delta x=\frac{5k}{n}[/tex][tex]f(x_k)=f(\frac{5k}{n})=(\frac{5k}{n})^2+5=\frac{25k^2}{n^2}+5[/tex]Now Area=
[tex]\begin{gathered} \lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{5}{n}(\frac{25k^2}{n^2}+5) \\ =\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{125k^2}{n^3}+\frac{25}{n} \\ =\lim_{n\to\infty}(\frac{125}{n^3}\sum_{k\mathop{=}1}^nk^2+\frac{25}{n}\sum_{k\mathop{=}1}^n1) \\ =\lim_{n\to\infty}(\frac{125}{n^3}.\frac{1}{6}n(n+1)(2n+1)+\frac{25}{n}n) \\ =\lim_{n\to\infty}(\frac{125(n+1)(2n+1)}{6n^2}+25) \\ =\lim_{n\to\infty}(\frac{125}{6}(1+\frac{1}{n})(2+\frac{1}{n})+25) \\ =\frac{125}{6}\times2+25=66.6 \end{gathered}[/tex]So the required area is 66.6 sq units.
