How do you integrate $\int t \sqrt[3]{t - 4} d t$ $int troot3(t-4)dt$ ?

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How do you integrate $\int t \sqrt[3]{t - 4} d t$ $int troot3(t-4)dt$ ?

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Answer 1 · 2017-01-29T23:47:30

$\frac{3}{7} {(t - 4)}^{\frac{7}{3}} + 3 {(t - 4)}^{\frac{4}{3}} + C$ $3/7(t-4)^(7/3)+3(t-4)^(4/3)+C$

Explanation:

$I = \int t \sqrt[3]{t - 4} . d t$ $I=inttroot3(t-4)color(white).dt$

Apply the substitution $u = t - 4$ $u=t-4$ . This also implies that $t = u + 4$ $t=u+4$ and $d u = d t$ $du=dt$ . Then:

$I = \int (u + 4) \sqrt[3]{u} . d u$ $I=int(u+4)root3ucolor(white).du$

We can write $\sqrt[3]{u}$ $root3u$ with a fractional exponent and then distribute:

$I = \int (u + 4) u^{\frac{1}{3}} . d u$ $I=int(u+4)u^(1/3)color(white).du$

$I = \int u^{\frac{4}{3}} . d u + 4 \int u^{\frac{1}{3}} . d u$ $I=intu^(4/3)color(white).du+4intu^(1/3)color(white).du$

Integrate both using the rule $\int u^{n} . d u = \frac{u^{n + 1}}{n + 1} + C$ $intu^ncolor(white).du=u^(n+1)/(n+1)+C$ :

$I = \frac{u^{\frac{7}{3}}}{\frac{7}{3}} + 4 (\frac{u^{\frac{4}{3}}}{\frac{4}{3}}) + C$ $I=u^(7/3)/(7/3)+4(u^(4/3)/(4/3))+C$

$I = \frac{3}{7} u^{\frac{7}{3}} + 3 u^{\frac{4}{3}} + C$ $I=3/7u^(7/3)+3u^(4/3)+C$

Since $u = t - 4$ $u=t-4$ :

$I = \frac{3}{7} {(t - 4)}^{\frac{7}{3}} + 3 {(t - 4)}^{\frac{4}{3}} + C$ $I=3/7(t-4)^(7/3)+3(t-4)^(4/3)+C$

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How do you integrate $\int t \sqrt[3]{t - 4} d t$ $int troot3(t-4)dt$ ?

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How do you integrate $\int t \sqrt[3]{t - 4} d t$ $int troot3(t-4)dt$ ?