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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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Try plotting the curves to get a feel for what the question asks you to do. – Srivatsan Oct 24 '11 at 14:46
How many solutions do you find to the pair of equations? Most you will not be able to find algebraically. – Ross Millikan Oct 24 '11 at 17:08
WolframAlpha shows five places where the equations coincide, and four bits of area to be considered. This is not an obvious calculation. – Peter Phipps Oct 24 '11 at 23:19
In view of the other comments, I'd advise checking to make sure that the problem is being quoted correctly. – Gerry Myerson Oct 25 '11 at 5:43
I'm still not convinced the problem is quoted correctly. – Gerry Myerson Oct 25 '11 at 8:21
up vote 2 down vote accepted

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

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


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

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

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

The diagram below may help you.

enter image description here

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What about between 2.142 and $\pi$? – Gerry Myerson Oct 25 '11 at 8:20
@GerryMyerson: Thanks for your correct observation. I have edited the post. – NoChance Oct 25 '11 at 8:49

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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Step 1 is the hard part, for this problem. – Gerry Myerson Oct 25 '11 at 8:19
For this particular problem you're not going to find exact solutions; use a graphing calculator (or Wolfram Alpha) to find the approximate crossings. – user7530 Oct 25 '11 at 8:24


  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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Step 1 is the hard part, for this problem. – Gerry Myerson Oct 25 '11 at 8:19
@GerryMyerson No it is not. Look at $f(x)=2\sin x - 1 -x/\pi$, it is easy to see that $f$ has at least 2 zeros (since $f(0)<0, f(\pi/2)>0, f(\pi)<0$). Also, $f'(x)=2\cos x -1/\pi$ which has exactly one zero in $[0,\pi]$. Now suppose you had more than two zeros in the interval - the mean value theorem would then give more at least two zeros for $f'$. – AD. Oct 25 '11 at 11:52
@GerryMyerson The above would at least give an expression for the area. – AD. Oct 25 '11 at 11:54
I'm sorry, I did not read carefully. What I meant to point out was that in this problem finding the actual intersection points is the hard part. So it's really step 3, where you need the values of the $x_i$. – Gerry Myerson Oct 25 '11 at 12:04
@GerryMyerson I agree, I also agree to your last comment to the problem - there might be an error in the quote of the homework. – AD. Oct 25 '11 at 13:25

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