1. ## integrals

Hi everyone,

Could you please tell me if the antiderivative of e^2x is e^2x/2 and if 2e^2x is e^2x/2? I got both of these answers by doing the opposite of taking the derivative. For example, the derivative of e^2x is e^2x(2), so wouldn't the antiderivative be e^2x/2?

Thank you very much

2. Originally Posted by chocolatelover
Hi everyone,

Could you please tell me if the antiderivative of e^2x is e^2x/2 and if 2e^2x is e^2x/2? I got both of these answers by doing the opposite of taking the derivative. For example, the derivative of e^2x is e^2x(2), so wouldn't the antiderivative be e^2x/2?

Thank you very much
how can the antiderivative of $\frac 12e^{2x}$ be the same as that of $2e^{2x}$? won't they differ by a factor of 4?

$\int e^{2x}~dx = \frac 12 e^{2x} + C$

we can get this, as you say, by reversing the differentiation operation. or the more standard way to do it is by a substitution of $u = 2x$. in addition, it is simple enough (if memorizing something is simple for me, it means it's simple, my memory is horrible) to memorize $\int e^{kx}~dx = \frac 1k e^{kx} + C$, where $k$ is a constant, and $k \ne 0$

now what do you think $\int 2e^{2x}~dx$ is?

3. You can also memorize the rule that if $F(x) +C= \int f(x) dx$ then $\int f(ax+b) dx = \frac{1}{a}F(ax+B)+C$ if $a\not = 0$.

4. Originally Posted by ThePerfectHacker
You can also memorize the rule that if $F(x) +C= \int f(x) dx$ then $\int f(ax+b) dx = \frac{1}{a}F(ax+B)+C$ if $a\not = 0$.
You must be forgetting about linear functions. Why would you divide by the constant a in that case?

5. What are you taking about?

6. Originally Posted by colby2152
You must be forgetting about linear functions. Why would you divide by the constant a in that case?
he is talking about linear functions. his conclusion follows by integration by substitution. that's how i do it as well. whenever i have a composite function in which the inner function is linear, i don't go through the trouble of integrating by substitution, i just know the integral will be the integral of the whole function divided by the derivative of the inner function--pretty much the direct opposite of the chain rule for derivatives (which is what the method of substitution models in general)

7. Originally Posted by ThePerfectHacker
Your rule fails if you let f(x) = x.

8. Originally Posted by colby2152
Your rule fails if you let f(x) = x.
How?

$\int x dx = \frac{1}{1}F(1x+0)+C$ where $F(x) = \int xdx = \frac{1}{2}x^2$.

9. Originally Posted by ThePerfectHacker
How?

$\int x dx = \frac{1}{1}F(1x+0)+C$ where $F(x) = \int xdx = \frac{1}{2}x^2$.
Let a = 2, so $F(2x) = 2x^2$... so unless I am mistaken your rule would give an answer of just $x^2$

10. Originally Posted by colby2152
Let a = 2, so $F(2x) = 2x^2$... so unless I am mistaken your rule would give an answer of just $x^2$
You are mistaken.
$\int (2x) dx = \frac{1}{2}F(2x)+C$ where $F(x) = \int x dx = \frac{1}{2}x^2$.
Thus,
$\frac{1}{2}F(2x) = \frac{1}{2}\cdot \frac{1}{2}(2x)^2 = x^2$