calculus integration?

Question

integration help

integral of: 2xe^(10x)dx

Answer 1

Integration by parts works, but you should really learn tabular integration by parts. It will make your life much simpler.

f(x): take the derivative and switch the sign each time
g(x): take the integral
then simply sum all of the products as shown below.

f(x) = 2x
------------
1. 2x

2. -2

3. 0

g(x) = e^(10x)
------------------
1. e^(10x)

2.(1/10)e^(10x)

3.(1/100)e^(10x)

The solution is (f(x)1 * g(x)2) + (f(x)2 * g(x)3)

=

(x/5)*e^(10x) - (1/50)*e^(10x)

It may seem a little confusing at first, but tabular integration by parts will save you a lot of headaches from all of the algebra involved with the uvvdu stuff.

Here is a link to explain it better

http://www.maa.org/pubs/Calc_articles/ma035.pdf

hope this helps.

Answer 2

Let I = integration of {2xe^(10x)dx}
Integrating by parts
I = 2x{integration {e^(10x)dx} - integration{{integration {e^(10x)dx}{differentiation of (2x)
I = (2x/10) {e^(10x)dx} - integration{(1/10) {e^(10x)dx}(2)
I = (2x/10) {e^(10x)dx - integration{(1/10) {e^(10x)dx}(2)
I =(x/5){e^(10x)} - (1/50)){e^(10x)} +C
I ={e^(10x)} {(10x - 1)/50} +C

Answer 3

Use part integral

u = x and v' = 10 e^(10x)

int uv' = uv - int u'v

u' is 1 and v is e^(10x)

Since you have only a 2, multiply and divide the integral by 5

Use the fact that

Int e^(10x) = [e^(10x)]/10

Ana

Answer 4

If you follow Anas' advice (and it's very good advice ☺) you should get

((10x - 1)e^(10x))/50 + C


Doug

Answer 5

?sin(5x)cos(7x) dx word that sin(12x) = sin(5x)cos(7x) + cos(5x)sin(7x) and sin(-2x) = sin(5x)cos(7x) - cos(5x)sin(7x) Now including both equations: sin(12x) - sin(2x) = 2sin(5x)cos(7x) => sin(5x)cos(7x) = a million/2 * [sin(12x) - sin(2x)] a million/2 * ?sin(12x) - sin(2x) dx a million/2 * [-a million/12*cos(12x) + a million/2*cos(2x)] + C