# Find indefinite integral. \int \frac{\sqrt{t}}{2}dt

Find indefinite integral.
$\int \frac{\sqrt{t}}{2}dt$
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Steve Hirano
Step 1
Consider the indefinite integral
$\int \frac{\sqrt{t}}{2}dt$
In order to evaluate this indefinite integral, use the power rule of integration
$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}$
Step 2
Using the exponent property $\sqrt{n}\left\{a\right\}={a}^{\frac{1}{n}}$, the given integral can be written as
$\int \frac{\sqrt{t}}{2}dt=\frac{1}{2}\int \sqrt{t}dt$
$=\frac{1}{2}\int {t}^{\frac{1}{2}}dt$
Now using the power rule of integration, we get
$\int \frac{\sqrt{t}}{2}dt=\frac{1}{2}\int {t}^{\frac{1}{2}}dt$
$=\frac{1}{2}\left[\frac{{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]+C$
$=\frac{1}{2}\left[\frac{{t}^{\frac{1+2}{2}}}{\frac{1+2}{2}}\right]+C$
$=\frac{1}{2}\left[\frac{{t}^{\frac{3}{2}}}{\frac{3}{2}}\right]+C$
$=\frac{1}{2}\left[\frac{2{t}^{\frac{3}{2}}}{3}\right]+C$
$=\frac{{t}^{\frac{3}{2}}}{3}+C$
Thus, the indefinite integral is $\int \frac{\sqrt{t}}{2}dt=\frac{{t}^{\frac{3}{2}}}{3}+C$ or $\int \frac{\sqrt{t}}{2}dt=\frac{t\sqrt{t}}{3}+C$
Here C is the integration constant.

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levurdondishav4
Given:
It is required to calculate:
$\int \frac{\sqrt{t}}{2}dt$
Lets

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nick1337

$\int \frac{\sqrt{t}}{2}dt$
$\frac{1}{2}×\int \sqrt{t}dt$
$\frac{1}{2}×\int {t}^{\frac{1}{2}}dt$
$\frac{1}{2}×\frac{2t\sqrt{t}}{3}$
$\frac{t\sqrt{t}}{3}$
$\frac{t\sqrt{t}}{3}+C$

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