Calculate each definite integral approximation in Exercises 23–40, and then find an error bound for your approximation. If it is possible to calculate the definite integral exactly, then do so and verify that the error bounds you found are accurate.

${\int }_{-1}^2\left(x^4-4x^3+4x^2\right)dx$,Simpson's rule,$n=8$

The given integral is${\int }_{-1}^2\left(x^4-4x^3+4x^2\right)dx$

Divide the given interval $[-1,2]$into 8 subparts to determine the width of the interval.

$$\begin{gathered} ∆x=\frac{x_8-x_0}{8} \\ =\frac{2-(-1)}{8} \\ =\frac{3}{8} \end{gathered}$$

Use this width of intervals to determine the points of interval.

$-1,-\frac{5}{8},-\frac{1}{4},\frac{1}{8},\frac{1}{2},\frac{7}{8},\frac{5}{4},\frac{13}{8},2$

Create a table of the values for the function at midpoint of intervals.

Use the above table values to determine the midpoint sum approximation.

$$\begin{gathered} SIMP\left(8\right)=\left(f\left(x_0\right)+4\left(f\right(x_1)+f(x_3)+f(x_5)+f(x_7)\right)+2\left(f\right(x_2)+f(x_4)+f(x_6\left)\right)+f\left(x_8\right)\left(\frac{∆x}{3}\right) \\ =\left(9+4\left(\frac{11025}{4096}+\frac{225}{4096}+\frac{3969}{4096}+\frac{1521}{4096}\right)+2\left(\frac{81}{256}+\frac{9}{16}+\frac{225}{256}\right)+0\right)\frac{1}{8} \\ =\left(9+\frac{16740}{1024}+\frac{450}{128}\right)\frac{1}{8} \\ =\frac{7389}{2048} \end{gathered}$$

Determine the fourth derivative of the given function.

$$\begin{gathered} f^1\left(x\right)=4x^3-12x^2+8x \\ {f}^2\left(x\right)=12x^2-24x+8 \\ {f}^3\left(x\right)=24x-24 \\ {f}^4\left(x\right)=24 \end{gathered}$$

Using the constant value of fourth derivative as value of M, the error of simpson's sum is determined as follows,

$$\begin{gathered} \left|E_{simp\left(n\right)}\right|\le \frac{M{(b-a)}^5}{180n^4} \\ \left|E_{simp\left(n\right)}\right|\le \frac{24(2-{(-1))}^5}{180{\left(8\right)}^4} \\ \left|E_{simp\left(n\right)}\right|\le 0.0079 \end{gathered}$$

Calculate the indefinite integral of the given function.

$$\begin{gathered} {\int }_{-1}^2\left(x^4-4x^3+4x^2\right)dx={\left[\frac{x^5}{5}-\frac{4x^4}{4}+\frac{4x^3}{3}\right]}_{-1}^2 \\ =\left[\left(\frac{2^5}{5}-2^4+\frac{4{\left(2\right)}^3}{3}\right)-\left(\frac{{(-1)}^5}{5}-{(-1)}^4+\frac{4{(-1)}^3}{3}\right)\right] \\ =\frac{33}{5}+(-15)+12 \\ =\frac{18}{5} \end{gathered}$$