Integration help (Graph Area)

f(x)= e^{-x}cosx 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.

Re: Integration help (Graph Area)

Quote:

Originally Posted by

**PedroB** f(x)= e^{-x}cosx 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

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.

Re: Integration help (Graph Area)

Quote:

Originally Posted by

**PedroB** f(x)= e^{-x}cosx 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

$\displaystyle B = \int\limits_0^{\pi /2} {{e^{ - x}}\cos x\,dx- \int\limits_{\pi /2}^{3\pi /2} {{e^{ - x}}\cos x\,dx} = A - \left. {\frac{{{e^{ - x}}}}{2}(\sin x - \cos x)} \right|_{\pi /2}^{3\pi /2} =\ldots= \frac{1+2e^{-\pi /2}+ e^{-3\pi /2}}{2}$

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.

Re: Integration help (Graph Area)