Given. ∫021+x24dx,n=8 (a) Use the Trapezoidal Rule to approximate the given integral with the specified...

Step 1
Given.
${\int }_0^2\sqrt[4]{1+x^2}dx,n=8$
For (a),
Trapezoidal rule:
We have that ${\displaystyle a=0,b=2,n=8}$
Therefore, ${\displaystyle \mathrm{\Delta }x=\frac{2-0}{8}-\frac{1}{4}}$
Devide the interval ${\displaystyle [0,2]\int oPSKn=8}$ subintervals of the length ${\displaystyle \mathrm{\Delta }x=\frac{1}{4}}$ with the following endpoints:
${\displaystyle a=0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1,\frac{5}{4},\frac{3}{2},\frac{7}{4},2=b}$
Now, just evaluate the function at these endpoints.
${\displaystyle f\left(x_0\right)=f\left(0\right)=1}$

$\begin{array}{}2f(x_1)=2f(\frac{1}{4})=\sqrt[4]{17}\approx 2.03054318486893\\ 2f(x_2)=2f(\frac{1}{2})=\sqrt{2}\sqrt[4]{5}\approx 2.11474252688113\\ 2f(x_3)=2f(\frac{3}{4})=\sqrt{5}\approx 2.23606797749979\\ 2f(x_4)=2f(1)=2\sqrt[4]{2}\approx 2.37841423000544\\ 2f(x_5)=2f(\frac{5}{4})=\sqrt[4]{41}\approx 2.53043953443524\\ 2f(x_6)=2f(\frac{3}{2})=\sqrt[4]{13}\sqrt{2}\approx 2.68534961428265\\ 2f(x_7)=2f(\frac{7}{4})=\sqrt[4]{65}\approx 2.83941151443368\\ f(x_8)=2f(2)=\sqrt[4]{5}\approx 1.49534878122122\end{array}$
Finally, just sum up the above values and multiply by ${\displaystyle \frac{\mathrm{\Delta }x}{2}=\frac{1}{8}}$
${\displaystyle \frac{1}{8}(1+2.03054318486893+2.11474252688113+2.23606797749979+2.37841423000544+2.53043953443524+2.68534961428265+2.83941151443368+1.49534878122122)=2.41378967045351}$
${\int }_0^2\sqrt[4]{1+x^2}dx\approx 2.413789$

Step 2
For (b),
Midpoint rule:
Divide the interval ${\displaystyle [0,2]}$ into ${\displaystyle n=8}$ subintervals of the length ${\displaystyle \mathrm{\Delta }x=\frac{1}{4}}$ with the following endpoints:
${\displaystyle a=0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1,\frac{5}{4},\frac{3}{2},\frac{7}{4},2=b}$
Now, just evaluate the function function at the midpoints of the subintervals.

$\begin{array}{}f(\frac{x_0+x_1}{2})=f(\frac{0+\frac{1}{4}}{2})=f(\frac{1}{8})=\frac{\sqrt{2}\sqrt[4]{65}}{4}=\approx 1.00388356821761\\ f(\frac{x_1+x_2}{2})=f(\frac{\frac{1}{4}+\frac{1}{2}}{2})=f(\frac{3}{8})=\frac{\sqrt{2}\sqrt[4]{73}}{4}=\approx 1.03344108112881\\ f(\frac{x_2+x_3}{2})=f(\frac{\frac{1}{2}+\frac{3}{4}}{2})=f(\frac{5}{8})=\frac{\sqrt{2}\sqrt[4]{89}}{4}=\approx 1.08593169283665\\ f(\frac{x_3+x_4}{2})=f(\frac{\frac{3}{4}+1}{2})=f(\frac{7}{8})=\frac{\sqrt[4]{113}\sqrt{2}}{4}=\approx 1.15272209425856\\ f(\frac{x_4+x_5}{2})=f(\frac{1+\frac{5}{4}}{2})=f(\frac{9}{8})=\frac{\sqrt[4]{145}\sqrt{2}}{4}=\approx 1.22686564967361\\ f(\frac{x_5+x_6}{2})=f(\frac{\frac{5}{4}+\frac{3}{2}}{2})=f(\frac{11}{8})=\frac{\sqrt[4]{185}\sqrt{2}}{4}=\approx 1.30391096842995\\ f(\frac{x_6+x_7}{2})=f(\frac{\frac{3}{2}+\frac{7}{4}}{2})=f(\frac{13}{8})=\frac{\sqrt{2}\sqrt[4]{233}}{4}=\approx 1.38131900381817\\ f(\frac{x_7+x_8}{2})=f(\frac{\frac{7}{4}+2}{2})=f(\frac{15}{8})=\frac{\sqrt{34}}{4}=\approx 1.45773797371133\end{array}$
Finally, just sum up the above values ans multiply by ${\displaystyle \mathrm{\Delta }x=\frac{1}{4}}$
${\displaystyle \frac{1}{4}(1.00388356821761+1.03344108112881+1.08593169283665+1.15272209425856+1.22686564967361+1.30391096842995+1.38131900381817+1.45773797371133)=2.41145300801867}$