How do you find the integral of $\int \frac{1}{{(x + 5)}^{\frac{1}{3}}} d x$ $int 1/(x+5)^(1/3) dx$ from 0 to 11?

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How do you find the integral of $\int \frac{1}{{(x + 5)}^{\frac{1}{3}}} d x$ $int 1/(x+5)^(1/3) dx$ from 0 to 11?

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Answer 1 · 2015-10-27T19:22:07

I found $5.14$ $5.14$

Explanation:

I would set $x + 5 = t$ $x+5=t$ so that $d x = d t$ $dx=dt$ and solve the indefinite integral first as:
$\int \frac{1}{t^{\frac{1}{3}}} d t = \int t^{- \frac{1}{3}} d t = \frac{t^{- \frac{1}{3} + 1}}{- \frac{1}{3} + 1} + c = \frac{t^{^\frac{2}{3}}}{\frac{2}{3}} + c = \frac{3}{2} t^{\frac{2}{3}} + c$ $int1/t^(1/3)dt=intt^(-1/3)dt=t^(-1/3+1)/(-1/3+1)+c=t^(^2/3)/(2/3)+c=3/2t^(2/3)+c$
but $t = x + 5$ $t=x+5$ so:
$= \frac{3}{2} {(x + 5)}^{\frac{2}{3}} + c$ $=3/2(x+5)^(2/3)+c$
Now we substitute the extrema and subtract:
$= \frac{3}{2} {(11 + 5)}^{\frac{2}{3}} - \frac{3}{2} {(0 + 5)}^{\frac{2}{3}} = 5.14$ $=3/2(11+5)^(2/3)-3/2(0+5)^(2/3)=5.14$

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How do you find the integral of $\int \frac{1}{{(x + 5)}^{\frac{1}{3}}} d x$ $int 1/(x+5)^(1/3) dx$ from 0 to 11?

1 Answer

Explanation:

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How do you find the integral of $\int \frac{1}{{(x + 5)}^{\frac{1}{3}}} d x$ $int 1/(x+5)^(1/3) dx$ from 0 to 11?