antiderivatives

Ry122

How do I find the antiderivative of
(12e^2x-5)
My attempt:
u=2x-5
u'=2
12u^u x 2
24e^2x-5
What do I do next?

alec_tronn

You started right, u = 2x-5, du = 2 dx. I'm not sure what you're trying to do in the next step. What you should try to do is 6du = 12 dx because you want to match the 12 in your original equation. With these subsitutions your new integrand is 6e^u.

VietDao29

Okay, the first step is correct.
We let u = 2x - 5
Then we find the differential of u, [tex]du = 2 dx \Rightarrow dx = \frac{du}{2}[/tex], now, with the substitution above, we'll change every x's in the expression into our newly-defined variable u:

[tex]\int 12 e ^ {2x - 5} dx = 12 \int e ^ u \left( \frac{du}{2} \right) = 6 \int e ^ u du = ...[/tex]
Can you go from here? After having its anti-derivative in terms of u, we should change u back to x, and complete the problem.
Is it clear? :)

HallsofIvy

It would be better to write 12e^(2x-5)dx. Including that "dx" reminds you that you need to replace IT as well. If u= 2x+5, then, yes, u'= 2 so du= 2dx.
You can either think "break that 12 into 6* 2 so 6e^(2x-5) (2dx) becomes 6e^u du" or write dx= (1/2) du so 6e^(2x-5)dx becomes 12e^u ((1/2)du)= 6e^u du again.

I suspect you were thinking of the chain rule when you multiplied by 2. That applies to differentiation. The anti-derivative is the opposite of the derivative so you divide by 2 rather than multiplying.

By the way- notice my use of parentheses. e^(2x-5) is NOT the same as e^2x- 5. Most people would interpret that latter as (e^(2x))- 5 and some might even interpret it as x(e^2)- 5. Use all the parentheses you need to make your meaning clear.