# Thread: Antiderivative

1. ## Antiderivative

I'm doing integrating by parts, and I came across a question that gave dv= er/2

Why is the anti-derivative 2er/2 ?

I know my work is wrong for what I have, and I'm really not sure why it is.

2. ## Re: Antiderivative

Because $\displaystyle \left (2e^\frac{r}{2} \right )'=2\cdot e^\frac{r}{2} \cdot \left ( \frac{r}{2} \right )'=2\cdot e^\frac{r}{2} \cdot \frac{1}{2}=e^\frac{r}{2}$.

3. ## Re: Antiderivative

thanks, is it a general rule that any e^r/x antiderivative is x^e/x ??

4. ## Re: Antiderivative

You're welcome! ^^

Uhm, I don't know what you actually wanted to ask, but yes, there is a rule:

$\displaystyle \left (e^{f(x)} \right )'=f'(x) \cdot e^{f(x)}$

Also, if $\displaystyle f(x)=ax+b$, then:

$\displaystyle \left (e^{ax+b} \right )'=a \cdot e^{ax+b}$

So the antiderivate of $\displaystyle e^{ax+b}$ is $\displaystyle \frac{1}{a}e^{ax+b}$.

5. ## Re: Antiderivative

Originally Posted by Steelers72
thanks, is it a general rule that any e^r/x antiderivative is x^e/x ??
Be careful about using "x" there since it will typically be interpreted as a variable and you mean it as a constant here. It is true that the derivative of e^(ar), with respect to r, where a is a constant, is ae^(ar) so that the derivative of e^(ar)/a, with respect to r, is ae^(ar)/a= e^(ar) so that the antiderivative of e^(ar) (again with constant a) is e^(ar)/a.

6. ## Re: Antiderivative

I see what you mean

so e^r/constant = constant*e^r/constant ?