So, came across the following question while I was studying for a test:

"Let $g$ be a differentiable function such that $\int g(x)e^{\frac{x}{4}}dx=4g(x)e^{\frac{x}{4}}-\int 8x^2e^{\frac{x}{4}}dx$ What's a possible expression for $g(x)$?"

We're talking about integration by parts, and I assumed that I should apply that formula ($\int udv = uv-\int vdu$) here, but it seems like after using that that there's no formula given for $g(x)$? I'm a bit lost and any help would be appreciated!

  • $\begingroup$ $ \frac{2}{3} x^3 $? $\endgroup$ – Adam Rubinson Oct 27 '20 at 1:52
  • $\begingroup$ Ok, I'm a bit confused on how you got there though. $\endgroup$ – Tyler Oct 27 '20 at 1:57
  • $\begingroup$ I would differentiate both sides and compare. $\endgroup$ – Kenta S Oct 27 '20 at 2:00
  • $\begingroup$ Ok, so I'm still a little confused, if I take the derivative of an integral that cancels it out right? So it'd basically be the same thing as just taking the integral signs away and going from there? $\endgroup$ – Tyler Oct 27 '20 at 2:03
  • $\begingroup$ IBP is: $ \int u v’ dx = u v - \int u’ v dx $. Here, my v’ is $e^{\frac{\pi}{4}}$ and then I figured out that $u$ is $g(x)$ and must equal $\frac{2}{3} x^3$. $\endgroup$ – Adam Rubinson Oct 27 '20 at 2:03

I write down the two terms so that you can compare them.

$$\int g(x)\cdot e^{\frac{x}{4}}\, dx=g(x)\cdot 4e^{\frac{x}{4}}-\int 8x^2e^{\frac{x}{4}}\, dx$$

$\begin{array}{} \qquad \qquad \qquad \qquad \qquad \quad \Huge\Updownarrow \large ➀ & \qquad \Huge\Updownarrow \large ➁ & \quad \ \Huge\Updownarrow \large ➂ \end{array}$

$$\int g(x)^{}\cdot h^{'}(x)\, dx=g(x)\cdot h^{}(x)-\int g(x)^{'}\cdot h^{}(x)\, dx$$

$\large ➀$ Identify $h^{'}(x)$

$\large ➁$ Check if $h^{}(x)=4e^{\frac{x}4}$

$\large ➂$ Extract $h(x)$ so that the remaining part is $g^{'}(x)$. Then integrate to obtain $g^{}(x)$.


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