Integration?

Question

Evaluate : e^(-2x)cos3x dx

with limits 0 to 1 .

Pls show step by step.

Answer 1

The key to this problem is to realize that after integrating by parts twice, you can obtain an expression for [0, 1]∫e^(-2x) cos (3x) dx in terms of a multiple of the same integral, and that after you have done this, simple algebra will suffice to obtain the integral itself. First, integrate by parts. Let u=e^(-2x), v=sin (3x)/3, du=-2e^(-2x) dx, dv=cos (3x) dx:

[0, 1]∫e^(-2x) cos (3x) dx = e^(-2x) sin (3x)/3 |[0, 1] + [0, 1]∫2/3 e^(-2x) sin (3x) dx

Evaluate at the limits:

[0, 1]∫e^(-2x) cos (3x) dx = sin 3/(3e²) + [0, 1]∫2/3 e^(-2x) sin (3x) dx

Integrate by parts again -- u = 2/3 e^(-2x), du = -4/3 e^(-2x) dx, v = -cos (3x)/3, dv = sin (3x) dx:

[0, 1]∫e^(-2x) cos (3x) dx = sin 3/(3e²) - 2/9 e^(-2x) cos (3x)|[0, 1] - 4/9 [0, 1]∫e^(-2x) cos (3x) dx

Evaluate at the limits:

[0, 1]∫e^(-2x) cos (3x) dx = sin 3/(3e²) - 2 cos 3/(9e²) + 2/9 - 4/9 [0, 1]∫e^(-2x) cos (3x) dx

Add 4/9 [0, 1]e^(-2x) cos (3x) dx to both sides:

13/9 [0, 1]∫e^(-2x) cos (3x) dx = sin 3/(3e²) - 2 cos 3/(9e²) + 2/9

Multiply both sides by 9/13:

[0, 1]∫e^(-2x) cos (3x) dx = (3 sin 3/e² - 2 cos 3/e² + 2)/13

And we are done.

Answer 2

Thank Goodness that you are not referimg to true integration
That of cultures. And differientaion. Calculate ?
In pure mathematics, though why limit the equation with limts of 0 to 1?
I know that is standard practice, but you will need to enlarge that in real life,
Is it an obssion to ask folks stuff that you already know the asswer to ?
I now just sketch a graph, which is all it boils down to, for example, trying to work out whether our local water mill will provide enough power to run the electrics.
Probaby, and all is well.
No limits to the integration, mathematically.
Your example should be straight-forward, but I would have to look up cosines, given the value of x.
Heck, it is easy, I like conversation, but there is no simple answer to your Q.
THe solution is in the understanding.
All the best,
Bob

Answer 3

With domain 0 to 1.

Answer 4

Hello

I will use Integration by parts to do this one:

u = e^(-2x)
du = -2e^(-2x)
v= (1/3) sin 3x
dv= cos3x dx

So we have (e^(-2x))(1/3) sin 3x) - Integral ((1/3) sin 3x) (-2e^(-2x))

Lets do Integration by parts again.

u= -2e^(-2x)
du = 4e^(-2x)
v = (-1/9) cos 3x
dv = (1/3) sin 3x dx

(e^(-2x))(1/3) sin 3x) - [ (-2e^(-2x))((-1/9) cos 3x)) - Integral of ((-1/9) cos 3x)(4e^(-2x))..... Looks like this will go forever.

So I will plug Evaluate : e^(-2x)cos3x dx
with limits 0 to 1 into my TI-84 to see what I get.

When I do Fnint(e^(-2x)cos3x, x, 0, 1) = .43222

Hope this helps

Answer 5

Ok, let's go on :

Integrate(e^-2x*cos3x)dx = M

We need to use the part by part integration :

Let's set :

u = e^-2x >>>>>>>du = -2*e^-3x*dx

dv = cos3x >>>> v = 1/3*sin(3x)

Now, the integration wil be :

u*v - integrate(v*du)

e^-2x*1/3*sin(3x) + 2/3*integrate(sin(3x)*e^-2x)dx = M

Now, let's integrate : integrate(sin(3x)*e^-2x)dx

Now, let's set : u = sin3x >>>> du = 1/3*cos3x*dx

dv = e^-2x >>>>> v = -1/2*e^-2x

u*v - integrate(v*du)

-1/2*sin(3x)*e^-2x + integrate(e^-2x*cos3x)dx = integrate(sin(3x)*e^-2x)dx

Let's remember : e^-2x*1/3*sin(3x) + 2/3*integrate(sin(3x)*e^-2x)dx = M

e^-2x*1/3*sin(3x) + 2/3*(-1/2*sin(3x)*e^-2x + integrate(e^-2x*cos3x)dx) = Integrate(e^-2x*cos3x)dx

Integrate(e^-2x*cos3x)dx = M

e^-2x*1/3*sin(3x) - 1/3*sin3x*e^-2x + 2/3*M = M

M = e^-2x*sin3x - sin3x*e^-2x

Now the limits : 0 to 1

M = e^-2*sin3 - sin3*e^-2

That's it
Hope that helps