Kapalci

2022-06-27

How to find the general antiderivative of $f\left(x\right)=x\left(6-x{\right)}^{2}$?
I want to find the general antiderivative of $f\left(x\right)=x\left(6-x{\right)}^{2}$. However, I keep getting it wrong.
I am new to antiderivatives, but I think the first thing I should do is differentiate.
$\frac{d}{dx}\left(6-x{\right)}^{2}=2\left(6-x\right)\cdot -1=-2\left(6-x\right)$
${f}^{\prime }\left(x\right)=\left[\left(6-x{\right)}^{2}\cdot 1\right]+\left[-2\left(6-x\right)\cdot x\right]=\left(6-x{\right)}^{2}-2x\left(6-x\right)$
According to the antiderivative power rule, when $n\ne -1$, $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$.
So it seems like $\int \left(6-x{\right)}^{2}dx=\frac{\left(6-x{\right)}^{3}}{3}+C$.
However, I can't find a rule that seems like it would work with $-2x\left(6-x\right)$. The closest thing that I can find is the "Multiplication by Constant Rule", $\int cf\left(x\right)dx=c\cdot \int \left(f\left(x\right)\right)dx$, but I'm not sure if I'm allowed to change $2x\left(6-x\right)$ into the form $2\left(6x-{x}^{2}\right)$.
$\int 2\left(6x-{x}^{2}\right)$
$=2\cdot \int \left(6x-{x}^{2}\right)$
$=2\left(6\int x-\int {x}^{2}\right)$
$=2\left(\frac{6{x}^{2}}{2}-\frac{{x}^{3}}{3}\right)+C$
But $\frac{\left(6-x{\right)}^{3}}{3}-\left(6{x}^{2}-\frac{2{x}^{3}}{3}\right)+C$ is incorrect.

assumintdz

Expert

Step 1
Another way to do it is to substitute $y=x-6$, so
$f\left(x\right)=6{y}^{2}+{y}^{3}=\frac{d}{dy}\left(2{y}^{3}+\frac{1}{4}{y}^{4}\right)=\frac{d}{dx}\frac{{y}^{3}\left(y+8\right)}{4}=\frac{d}{dx}\frac{\left(x-6{\right)}^{3}\left(x+2\right)}{4}.$
Step 2
Thus the general antiderivative is $\frac{\left(x-6{\right)}^{3}\left(x+2\right)}{4}+C$.

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