$s$

$\={\int }_0^4\sqrt{1\+{\left(\frac{\ⅆ}{\ⅆ x} x^{3\/2}\right)}^2}\mathit{\ⅆ}x$

$\={\int }_0^4\sqrt{1\+{\left(\frac{3}{2}{x}^{1\/2}\right)}^2}\mathit{\ⅆ}x$

$\={\int }_0^4\sqrt{1\+9 x\/4}\mathit{\ⅆ}x$

${\int }_0^4\sqrt{1\+9 x\/4}\mathit{\ⅆ}x$

$\=\frac{4}{9}{\int }_1^{10}{u}^{1\/2} \mathit{\ⅆ}u$

$\=\frac{4}{9} {\left[\frac{u^{3\/2}}{3\/2}\right]}_{u\=1}^{u\=10}$

$\=\frac{8}{27}\left({10}^{3\/2}-1\right)$

$\=\frac{8}{27} \left(10\sqrt{10}-1\right)$

$\doteq 9.073415289$

Application of the ArcLength command

Loading Student:-Calculus1

$\mathrm{ArcLength}\left(x^{3\/2}\,x\=0..4\right)$ = $\frac{80}{27}\sqrt{10}-\frac{8}{27}$

Stepwise evaluation of the arc-length integral

${\int }_0^4\sqrt{1\+9 x\/4}\mathit{\ⅆ}x$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Integration Steps}\\ & & {\int }_0^4\frac{\sqrt{4+9x}}{2}\ⅆx\\ \text{▫}& & \text{1. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int \frac{\sqrt{4+9x}}{2}\ⅆx\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(x\right)\ⅆx=C\left(\int f\left(x\right)\ⅆx\right)\\ & \text{◦}& \text{This means:}\\ & & \int \frac{\sqrt{4+9x}}{2}\ⅆx=\frac{\left(\int \sqrt{4+9x}\ⅆx\right)}{2}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{\left({\int }_0^4\sqrt{4+9x}\ⅆx\right)}{2}\\ \text{▫}& & \text{2. Apply a change of variables to rewrite the integral in terms of}u\\ & \text{◦}& \text{Let}\\ & & 4+9x=u^2\\ & \text{◦}& \text{Differentiate both sides}\\ & & \frac{\ⅆ}{\ⅆx}\left(4+9x\right)=\frac{\ⅆ}{\ⅆu}\left(u^2\right)\\ & \text{◦}& \text{Evaluate}\\ & & 9\mathrm{dx}=2u\mathrm{du}\\ & \text{◦}& \text{Substitute the values back into the original}\\ & & {\int }_0^4\sqrt{4+9x}\ⅆx={\int }_{\sqrt{4}}^{\sqrt{40}}\frac{2u^2}{9}\ⅆu\\ & & \text{This gives:}\\ & & \frac{\left({\int }_{\sqrt{4}}^{\sqrt{40}}\frac{2u^2}{9}\ⅆu\right)}{2}\\ \text{▫}& & \text{3. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int \frac{2u^2}{9}\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(u\right)\ⅆu=C\left(\int f\left(u\right)\ⅆu\right)\\ & \text{◦}& \text{This means:}\\ & & \int \frac{2u^2}{9}\ⅆu=\frac{2\left(\int u^2\ⅆu\right)}{9}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{\left({\int }_{\sqrt{4}}^{\sqrt{40}}{u}^2\ⅆu\right)}{9}\\ \text{▫}& & \text{4. Apply the}\mathbf{\text{power}}\text{rule to the term}\int u^2\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule, for n}\text{≠}\text{-1}\\ & & \int u^n\ⅆu=\frac{u^{n+1}}{n+1}\\ & \text{◦}& \text{This means:}\\ & & \int u^2\ⅆu=\frac{u^{2+1}}{2+1}\\ & \text{◦}& \text{So,}\\ & & \int u^2\ⅆu=\frac{u^3}{3}\\ & \text{◦}& \text{Apply limits of definite integral}\\ & & \genfrac{}{}{0ex}{}{\frac{u^3}{3}}{\phantom{u=\sqrt{40}}}|\genfrac{}{}{0ex}{}{\phantom{\frac{u^3}{3}}}{u=\sqrt{40}}-\left(\genfrac{}{}{0ex}{}{\frac{u^3}{3}}{\phantom{u=\sqrt{4}}}|\genfrac{}{}{0ex}{}{\phantom{\frac{u^3}{3}}}{u=\sqrt{4}}\right)\\ & & \text{We can rewrite the integral as:}\\ & & \frac{40\sqrt{40}}{27}-\frac{4\sqrt{4}}{27}\end{array}$

Application of the ArcLength command

$\mathrm{ArcLength}\left(x^{3\/2}\,x\=0..\stackrel{\^}{x}\right)$ = $-\frac{8}{27}\+\frac{1}{3}\sqrt{9\hat{x}\+4}\hat{x}\+\frac{4}{27}\sqrt{9\hat{x}\+4}$

Stepwise evaluation of the integral

${\int }_0^{\stackrel{\^}{x}}\sqrt{1\+9 x\/4}\ⅆx$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Integration Steps}\\ & & {\int }_0^{\hat{x}}\frac{\sqrt{4+9x}}{2}\ⅆx\\ \text{▫}& & \text{1. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int \frac{\sqrt{4+9x}}{2}\ⅆx\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(x\right)\ⅆx=C\left(\int f\left(x\right)\ⅆx\right)\\ & \text{◦}& \text{This means:}\\ & & \int \frac{\sqrt{4+9x}}{2}\ⅆx=\frac{\left(\int \sqrt{4+9x}\ⅆx\right)}{2}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{\left({\int }_0^{\hat{x}}\sqrt{4+9x}\ⅆx\right)}{2}\\ \text{▫}& & \text{2. Apply a change of variables to rewrite the integral in terms of}u\\ & \text{◦}& \text{Let}\\ & & 4+9x=u^2\\ & \text{◦}& \text{Differentiate both sides}\\ & & \frac{\ⅆ}{\ⅆx}\left(4+9x\right)=\frac{\ⅆ}{\ⅆu}\left(u^2\right)\\ & \text{◦}& \text{Evaluate}\\ & & 9\mathrm{dx}=2u\mathrm{du}\\ & \text{◦}& \text{Substitute the values back into the original}\\ & & {\int }_0^{\hat{x}}\sqrt{4+9x}\ⅆx={\int }_{x=0}^{x=\hat{x}}\frac{2u^2}{9}\ⅆu\\ & & \text{This gives:}\\ & & \frac{\left({\int }_{x=0}^{x=\hat{x}}\frac{2u^2}{9}\ⅆu\right)}{2}\\ \text{▫}& & \text{3. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int \frac{2u^2}{9}\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(u\right)\ⅆu=C\left(\int f\left(u\right)\ⅆu\right)\\ & \text{◦}& \text{This means:}\\ & & \int \frac{2u^2}{9}\ⅆu=\frac{2\left(\int u^2\ⅆu\right)}{9}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{\left({\int }_{x=0}^{x=\hat{x}}{u}^2\ⅆu\right)}{9}\\ \text{▫}& & \text{4. Apply the}\mathbf{\text{power}}\text{rule to the term}\int u^2\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule, for n}\text{≠}\text{-1}\\ & & \int u^n\ⅆu=\frac{u^{n+1}}{n+1}\\ & \text{◦}& \text{This means:}\\ & & \int u^2\ⅆu=\frac{u^{2+1}}{2+1}\\ & \text{◦}& \text{So,}\\ & & \int u^2\ⅆu=\frac{u^3}{3}\\ & \text{◦}& \text{Apply limits of definite integral}\\ & & \genfrac{}{}{0ex}{}{\frac{u^3}{3}}{\phantom{u=\left(x=\hat{x}\right)}}|\genfrac{}{}{0ex}{}{\phantom{\frac{u^3}{3}}}{u=\left(x=\hat{x}\right)}-\left(\genfrac{}{}{0ex}{}{\frac{u^3}{3}}{\phantom{u=\left(x=0\right)}}|\genfrac{}{}{0ex}{}{\phantom{\frac{u^3}{3}}}{u=\left(x=0\right)}\right)\\ & & \text{We can rewrite the integral as:}\\ & & \frac{\left(\genfrac{}{}{0ex}{}{\frac{u^3}{3}}{\phantom{x=0..\hat{x}}}|\genfrac{}{}{0ex}{}{\phantom{\frac{u^3}{3}}}{x=0..\hat{x}}\right)}{9}\\ \text{▫}& & \text{5. Revert change of variable}\\ & \text{◦}& \text{Variable we defined in step}2\\ & & 4+9x=u^2\\ & & \text{This gives:}\\ & & \frac{{\left(4+9\hat{x}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{27}-\frac{4\sqrt{4}}{27}\end{array}$