Question

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.

${\int }_0^2\sqrt{1+x^3}dx,rightsum,n=8$

Short answer

Right sum(8) = 3.501744

|E|<=0.5

Step-by-step solution

Given Information 

${\int }_0^2\sqrt{1+x^3}dx,rightsum,n=8$

 Definite Integral 

$$\begin{gathered} {\int }_0^2\sqrt{1+x^3}dx \\ \mathrm{Evaluating}\hspace{0.17em}\mathrm{using}\hspace{0.17em}\mathrm{numerical}\hspace{0.17em}\mathrm{calculation}\hspace{0.17em}\mathrm{methods}.\mathrm{Use}\mathrm{calculator} \\ {\int }_0^2\sqrt{1+{\mathrm{x}}^3}\mathrm{dx}=3.24130 \end{gathered}$$

Right sum 

$$\begin{gathered} \sum _{i=0}^{2}f\left(x_i\right)∆x=\left[f\left(x_1\right)+f\left(x_2\right).........+f\left(x_2\right)\right]∆x \\ ∆x=\frac{b-a}{8}=\frac{2-0}{8}=\frac{1}{4} \\ Endpointsare0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1,\frac{5}{4},\frac{3}{2},\frac{7}{4},2 \\ Sum=\frac{1}{4}\left[\sqrt{1+{\left(\frac{1}{4}\right)}^3}+\sqrt{1+{\left(\frac{1}{2}\right)}^3}+\sqrt{1+{\left(\frac{3}{4}\right)}^3}+\sqrt{1+{\left(1\right)}^3}+\sqrt{1+{\left(\frac{5}{4}\right)}^3}+\sqrt{1+{\left(\frac{3}{2}\right)}^3}+\sqrt{1+{\left(\frac{7}{4}\right)}^3}+\sqrt{1+{\left(2\right)}^3}\right] \\ Sum=3.501744 \end{gathered}$$

Error 

$$\begin{gathered} f\left(x\right)=\sqrt{1+x^3} \\ f\text{'}\left(x\right)=-\frac{3x^2}{\sqrt{1+x^3}} \\ f\text{'}\text{'}\left(x\right)=-\frac{1}{x^2} \\ f\text{'}\text{'}\left(x\right)ispositiveontheinterval[0,2] \\ f\left(x\right)isincrea\sin gonthegiveninterval \\ \left|E\right|\le \left|f\right(2)-f(0\left)\left[\frac{2-0}{8}\right] \\ \right|E|\le |\sqrt{1+2^3}-\sqrt{1+0^3}|\left[\frac{1}{4}\right] \\ \le 0.5 \end{gathered}$$