Kapalci

Answered

2022-06-27

How to find the general antiderivative of $f(x)=x(6-x{)}^{2}$?

I want to find the general antiderivative of $f(x)=x(6-x{)}^{2}$. However, I keep getting it wrong.

I am new to antiderivatives, but I think the first thing I should do is differentiate.

$\frac{d}{dx}(6-x{)}^{2}=2(6-x)\cdot -1=-2(6-x)$

${f}^{\prime}(x)=[(6-x{)}^{2}\cdot 1]+[-2(6-x)\cdot x]=(6-x{)}^{2}-2x(6-x)$

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 $\int (6-x{)}^{2}dx=\frac{(6-x{)}^{3}}{3}+C$.

However, I can't find a rule that seems like it would work with $-2x(6-x)$. The closest thing that I can find is the "Multiplication by Constant Rule", $\int cf(x)dx=c\cdot \int (f(x))dx$, but I'm not sure if I'm allowed to change $2x(6-x)$ into the form $2(6x-{x}^{2})$.

$\int 2(6x-{x}^{2})$

$=2\cdot \int (6x-{x}^{2})$

$=2(6\int x-\int {x}^{2})$

$=2(\frac{6{x}^{2}}{2}-\frac{{x}^{3}}{3})+C$

But $\frac{(6-x{)}^{3}}{3}-(6{x}^{2}-\frac{2{x}^{3}}{3})+C$ is incorrect.

I want to find the general antiderivative of $f(x)=x(6-x{)}^{2}$. However, I keep getting it wrong.

I am new to antiderivatives, but I think the first thing I should do is differentiate.

$\frac{d}{dx}(6-x{)}^{2}=2(6-x)\cdot -1=-2(6-x)$

${f}^{\prime}(x)=[(6-x{)}^{2}\cdot 1]+[-2(6-x)\cdot x]=(6-x{)}^{2}-2x(6-x)$

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 $\int (6-x{)}^{2}dx=\frac{(6-x{)}^{3}}{3}+C$.

However, I can't find a rule that seems like it would work with $-2x(6-x)$. The closest thing that I can find is the "Multiplication by Constant Rule", $\int cf(x)dx=c\cdot \int (f(x))dx$, but I'm not sure if I'm allowed to change $2x(6-x)$ into the form $2(6x-{x}^{2})$.

$\int 2(6x-{x}^{2})$

$=2\cdot \int (6x-{x}^{2})$

$=2(6\int x-\int {x}^{2})$

$=2(\frac{6{x}^{2}}{2}-\frac{{x}^{3}}{3})+C$

But $\frac{(6-x{)}^{3}}{3}-(6{x}^{2}-\frac{2{x}^{3}}{3})+C$ is incorrect.

Answer & Explanation

assumintdz

Expert

2022-06-28Added 22 answers

Step 1

Another way to do it is to substitute $y=x-6$, so

$f(x)=6{y}^{2}+{y}^{3}=\frac{d}{dy}(2{y}^{3}+\frac{1}{4}{y}^{4})=\frac{d}{dx}\frac{{y}^{3}(y+8)}{4}=\frac{d}{dx}\frac{(x-6{)}^{3}(x+2)}{4}.$

Step 2

Thus the general antiderivative is $\frac{(x-6{)}^{3}(x+2)}{4}+C$.

Another way to do it is to substitute $y=x-6$, so

$f(x)=6{y}^{2}+{y}^{3}=\frac{d}{dy}(2{y}^{3}+\frac{1}{4}{y}^{4})=\frac{d}{dx}\frac{{y}^{3}(y+8)}{4}=\frac{d}{dx}\frac{(x-6{)}^{3}(x+2)}{4}.$

Step 2

Thus the general antiderivative is $\frac{(x-6{)}^{3}(x+2)}{4}+C$.

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