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AP Calculus redo! I know the answer but can't figure out exactly how to get there. Thank you! I want to work through the steps. I don't just want the answer.

AP Calculus redo! I know the answer but can't figure out exactly how to get there-example-1
AP Calculus redo! I know the answer but can't figure out exactly how to get there-example-1
AP Calculus redo! I know the answer but can't figure out exactly how to get there-example-2
User Zipp
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You're approximating


\displaystyle\int_1^5 x^2\,\mathrm dx

with a Riemann sum, which comes in the form


\displaystyle\int_a^b f(x)\,\mathrm dx=\lim_(n\to\infty)\sum_(i=1)^nf(x_i)\Delta x_i

where
x_i are sample points chosen according to some decided-upon rule, and
\Delta x_i is the distance between adjacent sample points in the interval.

The simplest way of approximating the definite integral is by partitioning the interval into equally-spaced subintervals, in which case
\Delta x=\frac{b-a}n, and since
[a,b]=[1,5], we have


\Delta x=\frac{5-1}n=\frac4n

Using the right-endpoint method, we approximate the area under
f(x) with rectangles whose heights are determined by their right endpoints. These endpoints are chosen by successively adding the subinterval length to the starting point of the interval of integration.

So if we had
n=4 subintervals, we'd split up the interval of integration as


[1,5]=[1,2]\cup[2,3]\cup[3,4]\cup[4,5]

Note that the right endpoints follow a precise pattern of


2=1+\frac44

3=1+\frac84

4=1+\frac{12}4

5=1+\frac{16}4

The height of each rectangle is then given by the values above getting squared (since
f(x)=x^2). So continuing with the example of
n=4, the Riemann sum would be


\displaystyle\sum_(i=1)^4\left(1+\frac{4i}4\right)^2\frac44

For
n=5,


\displaystyle\sum_(i=1)^5\left(1+\frac{4i}5\right)^2\frac45

and so on, so that the definite integral is given exactly by the infinite sum


\displaystyle\lim_(n\to\infty)\sum_(i=1)^n\left(1+\frac{4i}n\right)^2\frac4n
User Will Huang
by
8.3k points
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