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[UPDATE in bold]

Find the area bounded by $y=2 \sin{x} - 1$ and $y = \frac{x}{\pi}$ for the range $0 \le x \le \pi$

I don't really know how to start... How can I find the intercepts of the curve/lines?

I get the 2 equations equals each other:

$ 2 \sin x - 1 = \frac{x}{\pi} $ $ 2 \sin x - 1 - \frac{x}{\pi} = 0 $ $ 2 \pi \sin x - \pi - x = 0 $

Then what do I do?

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    I'm still not convinced the problem is quoted correctly.2011-10-25

3 Answers 3

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Note: Post has been edited as per comment from GerryMyerson.

Area bound by 2 functions $f1$ and $f2$ can be found using integration of the function $f1-f2$

You must first identify the points where the 2 curves intersect in the are you are interested $0 \le x \le \pi$ in your case. This is what you started doing.

so you need to find values of x that satisfies:

$2 \pi \sin x - \pi - x = 0$

There are different ways you can do this depending on your background and tools available to you.

For example, you could either:

  1. Draw the curves

  2. Use numerical analysis

  3. Expand the function using series

  4. Use a software

Using 1,4 I found the points where the 2 curves intersect to have x values of: x1=0.647 and x2=2.142 (good for 2 decimal points)

Define $D(x) = 2 \sin x - 1 - \frac{x}{\pi}$

$A=ABS(A1) + ABS(A2) + ABS(A3)$

Where:

$A1=\int_{0}^{0.647} D(x)$

$A2=\int_{0.647}^{2.142} D(x)$

$A3=\int_{2.142}^{\pi} D(x)$

The diagram below may help you.

enter image description here

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    @GerryMyerson: Thanks for your correct observation. I have edited the post.2011-10-25
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The area between the curves $y= f(x)$ and $y=g(x)$, for the domain $a \leq x \leq b$, is given by $\int_a^b |f(x)-g(x)| dx.$

Here's the reasoning behind this formula: the area between curves $f(x)$ and $g(x)$ is the same as the area between $f(x)-g(x)$ and the $x$-axis. (Intuitively, if you think of the original area as a mound of dirt, you're allowing the dirt to fall until it is flush with the ground; doing so doesn't change the area of the dirt.)

The integral of $f(x)-g(x)$ gives you the signed area under the curve. To get the unsigned area, you have to stick in an absolute value.

So the procedure for solving this kind of problem is as follows:

  1. Determine whether $f$ and $g$ cross anywhere in the domain; in other words, find all $x$ in the domain for which $f(x)-g(x)=0$.
  2. Split the integral above into $n+1$ pieces, where $n$ is the number of crossing point you found in step 1, so that on each piece $f(x)-g(x)$ is either all positive, or all negative.
  3. On each piece, replace the absolute value with $+$ or $-$, as appropriate.
  4. Integrate up each piece and add together the results to get the final answer.

Let me know if anything is unclear, or if you need more help.

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    For this particular problem you're not going to find exact solutions; use a graphing calculator (or Wolfram Alpha) to find the approximate crossings.2011-10-25
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Hint

  1. Try to find the number of intersections between the curves.

  2. Call the points of intersection $x_1, x_2, x_3,\ldots,x_n$

  3. Try to calculate $\int_0^{x_1} f(x) - g(x)\, dx + \int_{x_1}^{x_2} g(x) - f(x)\, dx +\cdots+(-1)^{n}\int_{x_n}^{\pi} f(x) - g(x)\, dx$

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    @GerryMyerson I agree, I also agree to your last comment to the problem - there might be an error in the quote of the homework.2011-10-25