Question

Calculus question. Please complete and explain the following.

Answer 1

We're approximating the area under the graph of the function $f(x)=x^3+8$ over the interval [-2, 2] by

1. partitioning the integration interval into $n=4$ subintervals,
2. building rectangles whose lengths are equal to the length of the corresponding subinterval and whose heights are equal to the value of $f(r_i)$, where $r_i$ denotes the right endpoint of the $i$-th subinterval, and
3. computing the areas of each rectangle and adding these areas together.

Splitting [-2, 2] into 4 intervals gives

[-2, -1], [-1, 0], [0, 1], [1, 2]

Each subinterval has length 1. The right endpoints of the $i$-th subinterval, where $1\le i\le4$, are given by the (arithmetic) sequence

$r_i=-1+1(i-1)=i-2$

The area of the rectangle over the $i$-th subinterval is

$A_i=f(i-2)=(i-2)^3+8=i^3-6i^2+6i$

and so the definite integral is approximately

$\displaystyle\int_{-2}^2(x^3+8)\,\mathrm dx\approx\sum_{i=1}^4(i^3-6i^2+6i)$

There are well-known formulas for computing the sums of powers of consecutive (positive) integers. The ones we care about are

$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$

$\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4$

So we get

$\displaystyle\int_{-2}^2(x^3+8)\,\mathrm dx\approx40$