integration?

Question

how would i integrate e^(-2x) dx ?

Answer 1

intergral of e^(-2x) would be e^(-2x)/(-2)

Answer 2

Hello

The integration of e^x dx is e^x dx. But since -2x is the power you should use u substitution.

let u =-2x and du = -2 dx

So the final integration would be (-1/2)e^(-2x).

Double check this by taking its derivative and we get (-1/2)e^(-2x)(-2) = e^(-2x) dx. So it checks out.

Answer 3

There's a couple ways to do it. You could use u-substitution, you could try looking it up...there's a bunch of ways.

If you use u-sub, make u= -2x. Then du= -2dx or -du/2=dx. Substituting in, you now have:

S e^(u)*-du/2

(I'm using a capital "S" to denote the summation symbol here.)
Pull the -1/2 to the outside of the integral, and you now have:

-1/2 S e^(u) du

Integrating e^(u) just gives you e^(u) right back again (plus your annoying arbitrary constant C), so you wind up with:

-1/2 e^(u) +C

Substitute for u so you have all x's in the answer:

-1/2 e^(-2x) +C

And you're done!

Answer 4

Try putting -1/2 before the bracket and -2 within the bracket. You now have -1/2( -2e^-2x dx. This integrates to -1/2e^-2x + c

Answer 5

Substitution is hard work and it is easy to make mistakes.

You really need to know the standard integral
∫e^(ax) dx = (1/a) e^(ax) + c and then it is easy as well as quick and reliable.

∫e^(-2x) dx = (1/-2) e^(-2x) + c

Answer 6

The usual way. Transform, transform, transform.
Here U= -2x and dU= -2dx
Then we have S (-1/2) exp(U) dU which integrates to -1/2 exp(U)+c1 . Then we have
the integral in terms of x as (-1/2) exp(-2x) + c1

Answer 7

do 1/-2 and put it at the front and multiply by e^-(2x) and add on the constant of integration (+C)

In this case:

-0.5e^-(2x)+C

Answer 8

integral e^(kx) = (1/k)e^(kx) + C, so...

(-1/2)e^(-2x) + C.