# Area bounded by three functions

How would I do the following question.

Sketch the region bounded by the graphs of

• $f(x)=cos(x)$
• $g(x)=x-\frac{\pi}{2}$ and
• $x=0$

I know that $\cos(x)=0$ has an intercept at $\frac{\pi}{2}$ and $g(x)=x-\frac{\pi}{2}=0$ has an intercept at $\frac{\pi}{2}$ as well.

So would I use $\quad\displaystyle \int_0^\frac{\pi}{2}\cos(x)-(x-\frac{\pi}{2})(dx)$

which is $\quad\displaystyle G(x)=\sin(x)-\frac{x^2}{2}-\frac{\pi x}{2}$

or is there something I am doing wrong?

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You are integrating between $x = 0$ and $\pi/2$; those are the correct bounds of integration. One bound is the y-axis ($x = 0$), and the other bound is at $x = \pi/2$ where the functions $f(x) = g(x)$ intersect. And the upper curve on that interval is $\cos x$. All good, so far:

Simple algebraic mis-step: Recall that $a - (b - c) = a - b + c$, so...

$$\int_0^{\pi/2} \cos x - \left(x - \frac{\pi}{2}\right)\, dx = \int_0^{\pi/2} \left(\cos x - x + \frac{\pi}{2}\right) \,dx$$

And so, evaluating the integral gives us $$I(x) = \sin x - \dfrac{x^2}{2} + \dfrac{\pi x}{2}$$

And you need to then evaluate $I(\pi/2) - I(0)$.

To check your work: $$I(\pi/2) = \left(1 - \dfrac{\pi^2}{8} + \dfrac{\pi^2}{4}\right)\quad \text{and}\quad I(0) = 0 \quad \iff \quad \text{Area}\;= 1 + \frac{\pi^2}{8}$$

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I see that makes sense thanks. –  Fernando Martinez Apr 13 '13 at 22:09
You're welcome, Fernando. –  amWhy Apr 13 '13 at 22:12
Hmm for my area I got $G(X)=\frac{1}{2}+\frac{\pi^2}{4}$ –  Fernando Martinez Apr 13 '13 at 22:12
@amWhy: Very nice - graphs and all! +1 –  Amzoti Apr 14 '13 at 0:49
Thanks, amzoti! Graphs reveal a lot...even if only for confirmation! –  amWhy Apr 14 '13 at 0:50