Chapter 5.2, Problem 10E

Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places.

${\displaystyle {\int }_0^1\sqrt{x^3+1}dx},n=5$

To determine

To evaluate: The integral using Midpoint Rule.

Answer to Problem 10E

The value of ${\displaystyle {\int }_0^1\sqrt{\left(x^3+1\right)}dx}$ is 1.1097.

Explanation of Solution

Given information:

The integral function ${\displaystyle {\int }_0^1\sqrt{\left(x^3+1\right)}dx},n=5$.

Apply Midpoint Rule.

$\begin{array}{c}{\displaystyle {\int }_a^{b}f\left(x\right)dx}\approx {\displaystyle \sum _{i=1}^{n}f\left({\overline{x}}_i\right)\Delta x}\\ =\Delta x\left[f\left({\overline{x}}_1\right)+\mathrm{...}+f\left({\overline{x}}_n\right)\right]\end{array}$ (1)

Find the width $\left(\Delta x\right)$ using the relation:

$\Delta x=\frac{b-a}{n}$

Substitute 1 for b, 0 for a, and 5 for $n$ in Equation (1).

$\begin{array}{c}\Delta x=\frac{1-0}{5}\\ =\frac{1}{5}\\ =0.2\end{array}$

Calculate right end points $x_i$ using the relation:

$x_i=a+i\Delta x$ (2)

Calculate left end points $x_{i-1}$ using Equation (2)

Substitute $\left(i-1\right)$ for $i$ in Equation (2)

$x_{i-1}=a+\left(i-1\right)\Delta x$ (3)

Calculate mid points using the relation:

${\overline{x}}_i=\frac{1}{2}\left(x_{i-1}+x_i\right)$

Substitute $a+i\Delta x$ for $x_i$ and $\left[a+\left(i-1\right)\Delta x\right]$ for $x_{i-1}$

$\begin{array}{c}{\overline{x}}_i=\frac{1}{2}\left(x_{i-1}+x_i\right)\\ =\frac{1}{2}\left[a+\left(i-1\right)\Delta x+a+i\Delta x\right]\\ =\frac{1}{2}\left(a+i\Delta x-\Delta x+a+i\Delta x\right)\\ =\frac{1}{2}\times \left[2\left(a+i\Delta x\right)-\Delta x\right]\end{array}$

${\overline{x}}_i=\left(a+i\Delta x\right)-\frac{\Delta x}{2}$ (4)

Calculate ${\overline{x}}_1$ using Equation (4)

Substitute 0 for $a$, 1 for $i$,and $0.2$ for $\Delta x$ in Equation (4).

$\begin{array}{c}{\overline{x}}_i=\left(a+i\Delta x\right)-\frac{\Delta x}{2}\\ =\left(0+1\times 0.2\right)-\frac{1}{2}\times \left(0.2\right)\\ =0.2-0.1\\ =0.1\end{array}$

Calculate ${\overline{x}}_2$ using Equation (4).

Substitute 0 for $a$, 2 for $i$,and $0.2$ for $\Delta x$ in Equation (4).

$\begin{array}{c}{\overline{x}}_i=\left(a+i\Delta x\right)-\frac{\Delta x}{2}\\ =(0+2\times 0.2)-\frac{0.2}{2}\\ =0.4-0.1\\ =0.3\end{array}$

Calculate ${\overline{x}}_3$ using Equation (4)

Substitute 0 for $a$, 3 for $i$,and $0.2$ for $\Delta x$ in Equation (4)

$\begin{array}{c}{\overline{x}}_i=\left(a+i\Delta x\right)-\frac{\Delta x}{2}\\ =(0+3\times 0.2)-\frac{0.2}{2}\\ =0.6-0.1\\ =0.5\end{array}$

Calculate ${\overline{x}}_4$ using Equation (4)

Substitute 0 for $a$, 4 for $i$,and $0.2$ for $\Delta x$ in Equation (4)

$\begin{array}{c}{\overline{x}}_i=\left(a+i\Delta x\right)-\frac{\Delta x}{2}\\ =(0+4\times 0.2)-\frac{0.2}{2}\\ =0.8-0.1\\ =0.7\end{array}$

Calculate ${\overline{x}}_5$ using Equation (4)

Substitute 0 for $a$, 5 for $i$,and $0.2$ for $\Delta x$ in Equation (4)

$\begin{array}{c}{\overline{x}}_i=\left(a+i\Delta x\right)-\frac{\Delta x}{2}\\ =(0+5\times 0.2)-\frac{0.2}{2}\\ =1-0.1\\ =0.9\end{array}$

Compare the integral function ${\displaystyle {\int }_0^1\sqrt{\left(x^3+1\right)}dx}$ with Equation (1).

$f\left(x\right)=\sqrt{\left(x^3+1\right)}$

Calculate $f\left(x_i\right)$ using the Equation:

Substitute ${\overline{x}}_i$ for $x$.

$f\left({\overline{x}}_i\right)=\sqrt{\left({\overline{x}}_i^3+1\right)}$ (5)

Calculate $f\left({\overline{x}}_1\right)$ using Equation (5).

Substitute $0.1$ for ${\overline{x}}_1$ in the Equation (5)

$\begin{array}{c}f\left({\overline{x}}_i\right)=\sqrt{\left({\overline{x}}_i^3+1\right)}\\ =\sqrt{\left({\left(0.1\right)}^3+1\right)}\\ =\sqrt{1.001}\\ =1.0005\end{array}$

Calculate $f\left({\overline{x}}_2\right)$ using Equation (5).

Substitute $0.3$ for ${\overline{x}}_2$ in the Equation (5)

$\begin{array}{c}f\left({\overline{x}}_i\right)=\sqrt{\left({\overline{x}}_i^3+1\right)}\\ =\sqrt{\left({\left(0.3\right)}^3+1\right)}\\ =\sqrt{1.027}\\ =1.0134\end{array}$

Calculate $f\left({\overline{x}}_3\right)$ using Equation (5).

Substitute $0.5$ for ${\overline{x}}_3$ in the Equation (5)

$\begin{array}{c}f\left({\overline{x}}_i\right)=\sqrt{\left({\overline{x}}_i^3+1\right)}\\ =\sqrt{\left({\left(0.5\right)}^3+1\right)}\\ =\sqrt{1.125}\\ =1.0607\end{array}$

Calculate $f\left({\overline{x}}_4\right)$ using the Equation (5).

Substitute $0.7$ for ${\overline{x}}_4$ in the Equation (5)

$\begin{array}{c}f\left({\overline{x}}_i\right)=\sqrt{\left({\overline{x}}_i^3+1\right)}\\ =\sqrt{\left({\left(0.7\right)}^3+1\right)}\\ =\sqrt{1.343}\\ =1.1589\end{array}$

Calculate $f\left({\overline{x}}_5\right)$ using the Equation (5).

Substitute $0.9$ for ${\overline{x}}_4$ in the Equation (5)

$\begin{array}{c}f\left({\overline{x}}_i\right)=\sqrt{\left({\overline{x}}_i^3+1\right)}\\ =\sqrt{\left({\left(0.9\right)}^3+1\right)}\\ =\sqrt{1.729}\\ =1.3149\end{array}$

Calculate ${\displaystyle {\int }_0^1\sqrt{\left(x^3+1\right)}dx}$ using the Equation (1):

Substitute $\sqrt{\left(x^3+1\right)}$ for $f\left(x\right)$,$\sqrt{\left({\overline{x}}_i^3+1\right)}$ for $f\left({\overline{x}}_i\right)$, 0.2 for $\Delta x$,$1.0005$ for $f\left({\overline{x}}_1\right)$, $1.0134$ for $f\left({\overline{x}}_2\right)$, $1.0607$ for $f\left({\overline{x}}_3\right)$,and $1.1589$ for $f\left({\overline{x}}_4\right)$, $1.3149$ for $f\left({\overline{x}}_5\right)$, $0$ for $a$ and $1$ for $b$ in Equation (1).

$\begin{array}{c}{\displaystyle {\int }_0^1\sqrt{\left(x^3+1\right)}dx}=\Delta x\left[f\left({\overline{x}}_1\right)+f\left({\overline{x}}_2\right)+f\left({\overline{x}}_3\right)+f\left({\overline{x}}_4\right)+f\left({\overline{x}}_5\right)\right]\\ =0.2\left[1.0005+1.0134+1.0607+1.1589+1.3149\right]\\ =0.2\times 5.5484\\ =1.1097\end{array}$

Thus, the value of ${\displaystyle {\int }_0^1\sqrt{\left(x^3+1\right)}dx}$ is 1.1097.