Integration help (Graph Area)

• May 3rd 2012, 07:43 AM
PedroB
Integration help (Graph Area)
f(x)= e-xcosx with domain [0,3pi/2]

a) 'A' is the area between the x-axis, y-axis and f(x). 'B' is the area between the x-axis and f(x). Determine the ration of A to B (preferably in terms of 'e')

I was unable to do this question mainly because I couldn't find a way to integrate f(x), be it by u substitution or by parts. Any help would be greatly appreciated. Thanks in advance.
• May 3rd 2012, 07:55 AM
Plato
Re: Integration help (Graph Area)
Quote:

Originally Posted by PedroB
f(x)= e-xcosx with domain [0,3pi/2]

a) 'A' is the area between the x-axis, y-axis and f(x). 'B' is the area between the x-axis and f(x). Determine the ration of A to B (preferably in terms of 'e')

I was unable to do this question mainly because I couldn't find a way to integrate f(x), be it by u substitution or by parts. Any help would be greatly appreciated. Thanks in advance.

Use this result
• May 3rd 2012, 07:58 AM
PedroB
Re: Integration help (Graph Area)
The thing is I got this question in a non-calculator paper, I have to do everything manually, so I was looking for help on doing it like so. Thanks, though.
• May 3rd 2012, 02:47 PM
DeMath
Re: Integration help (Graph Area)
Quote:

Originally Posted by PedroB
f(x)= e-xcosx with domain [0,3pi/2]

a) 'A' is the area between the x-axis, y-axis and f(x). 'B' is the area between the x-axis and f(x). Determine the ration of A to B (preferably in terms of 'e')

I was unable to do this question mainly because I couldn't find a way to integrate f(x), be it by u substitution or by parts. Any help would be greatly appreciated. Thanks in advance.

Integrate by parts twice, then you'll get a linear algebraic equation with respect to the original integral.

\displaystyle \begin{aligned} I&= \int {{e^{ - x}}\cos x\,dx} = \int e^{-x}\,d(\sin x)= \\ &= {e^{ - x}}\sin x - \int {\sin x\,d({e^{ - x}})} = e^{-x}\sin x + \int {{e^{ - x}}\sin xdx} = \\ &= {e^{ - x}}\sin x - \int {{e^{ - x}}d(\cos x)} = e^{-x}\sin x - e^{-x}\cos x + \int {\cos x\,d({e^{ - x}})} = \\ &= {e^{ - x}}(\sin x - \cos x) - \int {{e^{ - x}}\cos x\,dx} = {e^{ - x}}(\sin x - \cos x) - I~~ \Leftrightarrow \\ &\Leftrightarrow~~ I + I = {e^{ - x}}(\sin x - \cos x)~~ \Leftrightarrow~~ I = \frac{e^{-x}}{2}(\sin x - \cos x) \end{aligned}

We find the intersection point of the graph with the x-axes (y=0)

$\displaystyle \left\{\begin{gathered}{e^{ - x}}\cos x = 0, \hfill \\x \in (0,3\pi /2) \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}\cos x = 0, \hfill \\x \in (0,3\pi /2) \hfill \\ \end{gathered} \right. \Leftrightarrow x = \frac{\pi}{2}$

$\displaystyle A = \int\limits_0^{\pi /2}e^{-x}\cos x\,dx= \ldots = \left. {\frac{e^{ - x}}{2}(\sin x - \cos x)} \right|_0^{\pi/2} =\ldots= \frac{1+e^{-\pi/2}}{2}$

Since the function takes negative values on the interval $\displaystyle \left(\frac{\pi}{2},\frac{3\pi}{2}\right)$​​, we have

$\displaystyle B = \int\limits_0^{\pi /2} {{e^{ - x}}\cos x\,dx- \int\limits_{\pi /2}^{3\pi /2} {{e^{ - x}}\cos x\,dx} = A - \left. {\frac{{{e^{ - x}}}}{2}(\sin x - \cos x)} \right|_{\pi /2}^{3\pi /2} =\ldots= \frac{1+2e^{-\pi /2}+ e^{-3\pi /2}}{2}$
• May 3rd 2012, 07:28 PM
CuriousMath
Re: Integration help (Graph Area)
Your suppose to use something called "integration by parts". Its derived from the product rule but you just need to know how to use formula. Its

Integration of uv' = uv - integration of u'v

Sorry not sure how to plug in formula still new to the site. You use this equation when you cant use u subsitution. In
Look up examples on youtube for an example like this.
• May 3rd 2012, 07:34 PM
DeMath
Re: Integration help (Graph Area)
This is the same.