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Here is the equation: $$\int t^2 \sqrt{t}dt$$

Here is my answer: $$\int t^2 \sqrt{t} = \dfrac12 \int t^{1/2}2t = \dfrac12 \dfrac{t^{3/2}}{3/2} = \dfrac12 \cdot \dfrac23 t^{3/2} + c$$

whereas here is the correct answer:

$$\dfrac27 \dfrac{t^{7}}2 + c$$

What am I doing wrong?

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  • $\begingroup$ You have $$\int t^{\left(2+\frac12\right)} dt=\int t^{\frac52}dt$$ right? $\endgroup$ Dec 7 '13 at 15:39
  • $\begingroup$ no i do not have that $\endgroup$
    – Cash Vai
    Dec 7 '13 at 15:42
  • $\begingroup$ then what do you have ? $\endgroup$ Dec 7 '13 at 15:43
  • $\begingroup$ I have what is above and I already notice my mistake. thank you! $\endgroup$
    – Cash Vai
    Dec 7 '13 at 15:45
  • $\begingroup$ @Cash Vai : are you doing integration by parts. $\endgroup$
    – Suraj M S
    Dec 7 '13 at 15:47
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$$\int t^2(t^{1/2}) dt= \int t^{(2 + 1/2)} \,dt= \int t^{\large \frac52}dt$$

Now use the power rule to integrate!

$$\int ax^{n} \,dx = \dfrac {ax^{n+1}}{n+1},\quad n\neq -1$$

Applying that here, noting that $\frac 52 + 1 = \frac 52 + \frac 22 = \frac 72$: $$\int t^{\large \frac52}dt = \frac{t^{\large \frac 72}}{\large \frac 72} + C = \frac 27 t^{\large \frac 72} + C $$

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  • $\begingroup$ Looks like you spotted it +1 $\endgroup$
    – Amzoti
    Dec 8 '13 at 2:51
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$$t^2\sqrt{t}=t^\frac{5}{2}$$ $$\int t^\frac{5}{2}dt=\frac{t^{\frac{5}{2}+1}}{\frac{5}{2}+1}+C$$

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$$\int t^2\sqrt{t} \ dt = \int t^2\cdot t^{\frac{1}{2}} \ dt = \int t^{\frac{5}{2}} \ dt = \frac{t^{\frac{5}{2}+1}}{\frac{5}{2}+1}+C=\frac{t^{\frac{7}{2}}}{\frac{7}{2}}+C =\frac{2}{7}t^{\frac{7}{2}}+C$$

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