<s>[INST] Find the function $f(x)$ whose graph passes through the point $(0,\frac{4}{3})$ and whose derivative is: $f'(x) = x \sqrt{16-x^2}$ Section 5.2
Can somebody verify this solution for me?
Find the function $f(x)$ whose graph passes through the point $(0,\frac{4}{3})$ and whose derivative is:
$f'(x) = x \sqrt{16-x^2}$

So the idea here is we integrate $f'(x)$ to get $f(x)=Something +C$ Where $C$ is our constant of integration, and then we use the fact that $f$ must pass through the point $(0,\frac{4}{3})$ to figure out what $C$ is.
So first lets integrate $f'(x) = x \sqrt{16-x^2}$
$\int x \sqrt{16-x^2}dx$
Let $u=16-x^2$. Then $\frac{du}{dx}=-2x$ and $\frac{du}{-2x}=dx$. And so we have:
$\int x \sqrt{16-x^2}dx$
$=\int x \sqrt{u}\frac{du}{-2x}$
$=\frac{-1}{2} \int \sqrt{u}du$
$=\frac{-1}{2} \frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C$
$=\frac{-1}{3} u^{\frac{3}{2}}+C$
$=\frac{-1}{3} (16-x^2)^{\frac{3}{2}}+C$
Okay. So by defintion $f(x)=\int f'(x)dx$. Thus $f(x) = \frac{-1}{3} (16-x^2)^{\frac{3}{2}}+C$. Lets plug in the point $(0,\frac{4}{3})$ and solve for $C$
$\frac{4}{3} = \frac{-1}{3} (16-0^2)^{\frac{3}{2}}+C$
$\rightarrow \frac{4}{3} = \frac{-1}{3} (16)^{\frac{3}{2}}+C$
$\rightarrow 4= -(16)^{\frac{3}{2}}$
$\rightarrow 4=-(4^3)+ \frac{C}{3}$
$\rightarrow 4=-64+ \frac{C}{3}$
$\rightarrow \frac{68}{3}=C$
Therefore, plugging this back into $f(x)$, we get:
$f(x) = \frac{-1}{3} (16-x^2)^{\frac{3}{2}}+\frac{68}{3}$
Unfortunately, the computer says the correct solution is:
$f(x) = \frac{-1}{3} (16-x^2)^{\frac{3}{2}}+\frac{68}{3}$
Can somebody find my error? Thanks!
 [/INST] The correction is : $\frac{4}{3}=\frac{-1}{3}(16)^{3/2} +C$ then $ 4=-1(16)^{3/2} +3C$
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