MHB What is the Chain Rule for Integration?

[karush]

$\tiny\text{Whitman 8.7.18 chain rule} $
$$\displaystyle
I=\int { \left({t}^{3/2}+47\right)^3 \sqrt{t} } \ d{t}
={ \left({t}^{3/2}+{47}^{}\right)^4/6 } + C$$
$$\begin{align}
\displaystyle
u& = {t}^{3/2}+47&
du&=\frac{3}{2}{t}^{1/2} \ d{t}& \\
\end{align}$$

$$I=\frac{3}{2}\int\left({u}\right)^3 du $$

Don't see the answer coming from this?

$\tiny\text
{from Surf the Nations math study group}$

 

[MarkFL]

Okay, let's take a look at your substitution:

$$u=t^{\frac{3}{2}}+47\,\therefore\,du=\frac{3}{2}t^{\frac{1}{2}}\,dt\implies dt=\frac{2}{3}t^{-\frac{1}{2}}\,du$$

So, we now have:

$$I=\frac{2}{3}\int u^3\,du$$

You only made a minor error in your substitution for $dt$...now you will get the stated result. :D

 

[karush]

$\tiny\text{Whitman 8.7.18 chain rule} $
$$\displaystyle
I=\int { \left({t}^{3/2}+47\right)^3 \sqrt{t} } \ d{t}
={ \left({t}^{3/2}+{47}^{}\right)^4/6 } + C$$
$$\begin{align}
\displaystyle
u& = {t}^{3/2}+47&
du&=\frac{3}{2}{t}^{1/2} \ d{t}& \\
\end{align}$$

$$I=\frac{2}{3}\int u^3 du
= \frac{2}{3}\cdot\frac{u^4}{4}
=\frac{u^4}{6}$$

Back substittute

$$I=\frac{\left({t}^{3/2}+47\right)^4 }{6}+C$$$\tiny\text
{from Surf the Nations math study group}$