Roland Manning

2022-07-01

Antiderivative of $8{t}^{-1/2}$
I am trying to find the antiderivative of $8{t}^{-1/2}$ but im just not understanding how to do this.
I saw someone get out of ${t}^{1/2}$ the answer $2{t}^{1/2}$. Can someone help me out?

Trey Ross

Step 1
Don't be worried about a negative exponent that's not -1: you apply the same rules as you would when integrating, say, $\phantom{\rule{thickmathspace}{0ex}}\int 8{t}^{3}dt,\phantom{\rule{thickmathspace}{0ex}}$, for example.
Those rules, recall, are:
Step 2
For any $n\ne -1$ and any number b,

$\text{So, since}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\left(-1/2\right)\ne -1,\phantom{\rule{1em}{0ex}}\int 8{t}^{-1/2}dt=\frac{8}{\left(\frac{1}{2}\right)}{t}^{\phantom{\rule{thinmathspace}{0ex}}\left(-\frac{1}{2}\right)\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}1\right)}+C\phantom{\rule{1em}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}\dots \phantom{\rule{thickmathspace}{0ex}}?$

Sonia Gay

Explanation:
$\int {x}^{\alpha }dx=\frac{1}{\alpha +1}{x}^{\alpha +1}+C$, as long as $\alpha \ne -1$. Also, note that $\int 8{t}^{-\frac{1}{2}}dt=8\int {t}^{-\frac{1}{2}}dt$.

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