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?

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

Beginner2022-07-02Added 30 answers

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,

$\int b\cdot {x}^{n}dx\phantom{\rule{1em}{0ex}}=\phantom{\rule{1em}{0ex}}b\int {x}^{n}dx\phantom{\rule{1em}{0ex}}=\phantom{\rule{1em}{0ex}}b\cdot \frac{1}{(n+1)}{x}^{(n+1)}+C\phantom{\rule{1em}{0ex}}\text{(where}C\text{is a constant)}.$

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

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,

$\int b\cdot {x}^{n}dx\phantom{\rule{1em}{0ex}}=\phantom{\rule{1em}{0ex}}b\int {x}^{n}dx\phantom{\rule{1em}{0ex}}=\phantom{\rule{1em}{0ex}}b\cdot \frac{1}{(n+1)}{x}^{(n+1)}+C\phantom{\rule{1em}{0ex}}\text{(where}C\text{is a constant)}.$

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

Sonia Gay

Beginner2022-07-03Added 7 answers

Explanation:

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

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

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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