Question

For the function given below, find a formula for the Riemann sum obtained by dividing the interval [0,5] into n equal subintervals and using the right-hand endpoint for each ck. then take a limit of this sum as n --> infinity to calculate the area under the curve over [0,5]. (the main problem is provided in the picture)

Answer 1

Given

we are given a function

`f(x)=x^2+5`

over the interval [0,5].

Required

we need to find formula for Riemann sum and calculate area under the curve over [0,5].

Step-by-step explanation

If we divide interval [a,b] into n equal intervals, then each subinterval has width

`\Delta x=(b-a)/(n)`

and the endpoints are given by

`a+k.\Delta x,\text{ for }0\leq k\leq n`

For k=0 and k=n, we get

`\begin{gathered} x_0=a+0((b-a)/(n))=a \\ x_n=a+n((b-a)/(n))=b \end{gathered}`

Each rectangle has width and height as

`\Delta x\text{ and }f(x_k)\text{ respectively.}`

we sum the areas of all rectangles then take the limit n tends to infinity to get area under the curve:

`Area=\lim_(n\to\infty)\sum_{k\mathop{=}1}^n\Delta x.f(x_k)`

Here

`f(x)=x^2+5\text{ over the interval \lbrack0,5\rbrack}`
`\Delta x=(5-0)/(n)=(5)/(n)`
`x_k=0+k.\Delta x=(5k)/(n)`
`f(x_k)=f((5k)/(n))=((5k)/(n))^2+5=(25k^2)/(n^2)+5`

Now Area=

`\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(5)/(n)((25k^2)/(n^2)+5) \\ =\lim_(n\to\infty)\sum_{k\mathop{=}1}^n(125k^2)/(n^3)+(25)/(n) \\ =\lim_(n\to\infty)((125)/(n^3)\sum_{k\mathop{=}1}^nk^2+(25)/(n)\sum_{k\mathop{=}1}^n1) \\ =\lim_(n\to\infty)((125)/(n^3).(1)/(6)n(n+1)(2n+1)+(25)/(n)n) \\ =\lim_(n\to\infty)((125(n+1)(2n+1))/(6n^2)+25) \\ =\lim_(n\to\infty)((125)/(6)(1+(1)/(n))(2+(1)/(n))+25) \\ =(125)/(6)*2+25=66.6 \end{gathered}`

So the required area is 66.6 sq units.