Wanda Kane

2021-12-28

Given.
${\int }_{0}^{2}\sqrt[4]{1+{x}^{2}}dx,n=8$
(a) Use the Trapezoidal Rule to approximate the given integral with the specified value of n. (Round your answer to six decimal places.)
___ (b) Use the Midpoint Rule to approximate the given integral with the specified value of n. (Round your answer to six decimal places.)
___ (c) Use Simpson's Rule to approximate the given integral with the specified value of n. (Round your answer to six decimal places.)

Alex Sheppard

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

$\begin{array}{}2f\left({x}_{1}\right)=2f\left(\frac{1}{4}\right)=\sqrt[4]{17}\approx 2.03054318486893\\ 2f\left({x}_{2}\right)=2f\left(\frac{1}{2}\right)=\sqrt{2}\sqrt[4]{5}\approx 2.11474252688113\\ 2f\left({x}_{3}\right)=2f\left(\frac{3}{4}\right)=\sqrt{5}\approx 2.23606797749979\\ 2f\left({x}_{4}\right)=2f\left(1\right)=2\sqrt[4]{2}\approx 2.37841423000544\\ 2f\left({x}_{5}\right)=2f\left(\frac{5}{4}\right)=\sqrt[4]{41}\approx 2.53043953443524\\ 2f\left({x}_{6}\right)=2f\left(\frac{3}{2}\right)=\sqrt[4]{13}\sqrt{2}\approx 2.68534961428265\\ 2f\left({x}_{7}\right)=2f\left(\frac{7}{4}\right)=\sqrt[4]{65}\approx 2.83941151443368\\ f\left({x}_{8}\right)=2f\left(2\right)=\sqrt[4]{5}\approx 1.49534878122122\end{array}$
Finally, just sum up the above values and multiply by $\frac{\mathrm{\Delta }x}{2}=\frac{1}{8}$
$\frac{1}{8}\left(1+2.03054318486893+2.11474252688113+2.23606797749979+2.37841423000544+2.53043953443524+2.68534961428265+2.83941151443368+1.49534878122122\right)=2.41378967045351$
${\int }_{0}^{2}\sqrt[4]{1+{x}^{2}}dx\approx 2.413789$

Gerald Lopez

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

$\begin{array}{}f\left(\frac{{x}_{0}+{x}_{1}}{2}\right)=f\left(\frac{0+\frac{1}{4}}{2}\right)=f\left(\frac{1}{8}\right)=\frac{\sqrt{2}\sqrt[4]{65}}{4}=\approx 1.00388356821761\\ f\left(\frac{{x}_{1}+{x}_{2}}{2}\right)=f\left(\frac{\frac{1}{4}+\frac{1}{2}}{2}\right)=f\left(\frac{3}{8}\right)=\frac{\sqrt{2}\sqrt[4]{73}}{4}=\approx 1.03344108112881\\ f\left(\frac{{x}_{2}+{x}_{3}}{2}\right)=f\left(\frac{\frac{1}{2}+\frac{3}{4}}{2}\right)=f\left(\frac{5}{8}\right)=\frac{\sqrt{2}\sqrt[4]{89}}{4}=\approx 1.08593169283665\\ f\left(\frac{{x}_{3}+{x}_{4}}{2}\right)=f\left(\frac{\frac{3}{4}+1}{2}\right)=f\left(\frac{7}{8}\right)=\frac{\sqrt[4]{113}\sqrt{2}}{4}=\approx 1.15272209425856\\ f\left(\frac{{x}_{4}+{x}_{5}}{2}\right)=f\left(\frac{1+\frac{5}{4}}{2}\right)=f\left(\frac{9}{8}\right)=\frac{\sqrt[4]{145}\sqrt{2}}{4}=\approx 1.22686564967361\\ f\left(\frac{{x}_{5}+{x}_{6}}{2}\right)=f\left(\frac{\frac{5}{4}+\frac{3}{2}}{2}\right)=f\left(\frac{11}{8}\right)=\frac{\sqrt[4]{185}\sqrt{2}}{4}=\approx 1.30391096842995\\ f\left(\frac{{x}_{6}+{x}_{7}}{2}\right)=f\left(\frac{\frac{3}{2}+\frac{7}{4}}{2}\right)=f\left(\frac{13}{8}\right)=\frac{\sqrt{2}\sqrt[4]{233}}{4}=\approx 1.38131900381817\\ f\left(\frac{{x}_{7}+{x}_{8}}{2}\right)=f\left(\frac{\frac{7}{4}+2}{2}\right)=f\left(\frac{15}{8}\right)=\frac{\sqrt{34}}{4}=\approx 1.45773797371133\end{array}$
Finally, just sum up the above values ans multiply by $\mathrm{\Delta }x=\frac{1}{4}$
$\frac{1}{4}\left(1.00388356821761+1.03344108112881+1.08593169283665+1.15272209425856+1.22686564967361+1.30391096842995+1.38131900381817+1.45773797371133\right)=2.41145300801867$

karton

Step 3
For (c),
Simpson's Rule:
Divide the interval [0, 2] into n=8 subintervals of the length $\mathrm{\Delta }x=\frac{1}{4}$ with the following endpoints:
$a=0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1,\frac{5}{4},\frac{3}{2},\frac{7}{4},2=b$
Now, just evaluate the function function at these endpoints.
$\begin{array}{}f\left({x}_{0}\right)=f\left(0\right)=1\\ 4f\left({x}_{1}\right)=4f\left(\frac{1}{4}\right)=2\sqrt[4]{17}\approx 4.06108636973786\\ 2f\left({x}_{2}\right)=2f\left(\frac{1}{2}\right)=\sqrt{2}\sqrt[4]{5}\approx 2.11474252688113\\ 4f\left({x}_{3}\right)=4f\left(\frac{3}{4}\right)=2\sqrt{5}\approx 4.47213595499958\\ 2f\left({x}_{4}\right)=2f\left(1\right)=2\sqrt[4]{2}\approx 2.37841423000544\\ 4f\left({x}_{5}\right)=4f\left(\frac{5}{4}\right)=2\sqrt[4]{41}\approx 5.06087906887049\\ 2f\left({x}_{6}\right)=2f\left(\frac{3}{2}\right)=\sqrt[4]{13}\sqrt{2}\approx 2.68534961428265\\ 4f\left({x}_{7}\right)=4f\left(\frac{7}{4}\right)=2\sqrt[4]{65}\approx 5.67882302886736\\ f\left({x}_{8}\right)=4f\left(2\right)=\sqrt[4]{5}\approx 1.49534878122122\end{array}$
Finally, just sum up the above values and multiply by $\frac{\mathrm{\Delta }x}{3}=\frac{1}{12}$
$\frac{1}{12}\left(1+4.06108636973786+2.11474252688113+4.47213595499958+2.37841423000544+5.06087906887049+2.68534961428265+5.67882302886736+1.49534878122122\right)=2.41223163123881$
${\int }_{0}^{2}\sqrt[4]{1+{x}^{2}}dx\approx 2.412232$

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