Find the function $f(x)$ whose graph passes through the point $(0,\frac{4}{3})$ and whose derivative is: $f'(x) = x \sqrt{16-x^2}$

Question

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:

$f'(x) = x \sqrt{16-x^2}$

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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!

Answer 1

Your mistake is introduced here:

$\rightarrow \frac{4}{3} = \frac{-1}{3} (16)^{\frac{3}{2}}+C$

$\rightarrow 4= -(16)^{\frac{3}{2}}+C$

What you did here is multiply by $3$ on both sides, I believe. You also went ahead and let the factor of $3$ be "absorbed" into the constant $C$ (after all, $C$ is a constant, and therefore so is $3C$). We do this a lot in more complicated integrals, so it's understandable you would do this.

However, the value of $C$ necessary to satisfy this problem is fixed - a very particular (indeed, just one) value of $C$ can work here, so we cannot let $3C$ be considered as just as nebulous a constant as $C$. It's like when you're solving $x-1=0$: if you have $3x$ instead, you'd get a very different solution.

But at least this means rectifying your solution is easy. From the second line forward, just replace $C$ with $3C$, and then an additional last step will let you conclude $C=68/3$.

Answer 2

The correction is : $\frac{4}{3}=\frac{-1}{3}(16)^{3/2} +C$ then $ 4=-1(16)^{3/2} +3C$