Mary Keefe

2021-12-28

Find the indefinite integral.
$\int x\sqrt{8-x}dx$

Bob Huerta

Step 1
To solve this indefinite integral, we have to use the substitution method.
Substitute 8-x = u, that implies -dx = du, and x = 8 - u.
By substituting these values in the above integral,
$=-\int \left(8-u\right)\sqrt{u}du$
$=-\int 8{u}^{\frac{1}{2}}-{u}^{\frac{3}{2}}du$
$=\frac{2}{5}{u}^{\frac{5}{2}}-\frac{16}{3}{u}^{\frac{3}{2}}+C$
Step 2
By resubstituting value u = (8 - x)
$=\frac{2}{5}{\left(8-x\right)}^{\frac{5}{2}}-\frac{16}{3}{\left(8-x\right)}^{\frac{3}{2}}+C$
This is the required integral.

Joseph Fair

$\int x\sqrt{8-x}dx$
$=\int \left({u}^{\frac{3}{2}}-8\sqrt{u}\right)du$
$=\int {u}^{\frac{3}{2}}du-8\int \sqrt{u}du$
$\int {u}^{\frac{3}{2}}du$
$=\frac{2{u}^{\frac{5}{2}}}{5}$
$\int \sqrt{u}du$
$=\frac{2{u}^{\frac{3}{2}}}{3}$
$\int {u}^{\frac{3}{2}}du-8\int \sqrt{u}du$
$=\frac{2{u}^{\frac{5}{2}}}{5}-\frac{16{u}^{\frac{3}{2}}}{3}$
$=\frac{2{\left(8-x\right)}^{\frac{5}{2}}}{5}-\frac{16{\left(8-x\right)}^{\frac{3}{2}}}{3}$
$\int x\sqrt{8-x}dx$
$=\frac{2{\left(8-x\right)}^{\frac{5}{2}}}{5}-\frac{16{\left(8-x\right)}^{\frac{3}{2}}}{3}+C$
$=\frac{2\left(-3x-16\right){\left(8-x\right)}^{\frac{3}{2}}}{15}+C$

karton

$\begin{array}{}x\sqrt{8-x}dx\\ \int t\sqrt{t}-8\sqrt{t}dt\\ \int t×{t}^{\frac{1}{2}}-8{t}^{\frac{1}{2}}dt\\ \int {t}^{\frac{3}{2}}-8{t}^{\frac{1}{2}}dt\\ \int {t}^{\frac{3}{2}}dt-\int 8{t}^{\frac{1}{2}}dt\\ \frac{2{t}^{2}\sqrt{t}}{5}-\frac{16t\sqrt{t}}{3}\\ \frac{2\left(8-x{\right)}^{2}×\sqrt{8-x}}{5}-\frac{16\left(8-x\right)\sqrt{8-x}}{3}\\ \frac{2\sqrt{8-x}\left(64-16x+{x}^{2}\right)}{5}-\frac{16\left(8-x\right)\sqrt{8-x}}{3}\end{array}$
$\frac{2\sqrt{8-x}\left(64-16x+{x}^{2}\right)}{5}-\frac{16\left(8-x\right)\sqrt{8-x}}{3}+C$