I do not understand the second task.

The first task is evaluating following integral, what I did and got $-2π$ $$ \int_0^{2\pi}x\sin(x)\,dx=-2\pi\approx-6.2832. $$

However the second question is this: Find the area between the curve $y=x \sin x$ and the $x$-axis, and the lines $x=0$ and $x=2π$.

Isn't this exactly the same as the first question, evaluating the definite integral?

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    $\begingroup$ No because is asking for the positive part of the graph i.e. $\max\{0,x\sin{x}\}$ $\endgroup$ – rlartiga Mar 19 '15 at 19:00
  • $\begingroup$ @rlartiga Not if it's interpreted the usual way, it's not. There would have to be a y=0 line for that. It is asking for something else then the first question though. $\endgroup$ – DRF Mar 19 '15 at 19:02
  • $\begingroup$ @DRF "and the x axis" is exactly $y=0$ $\endgroup$ – rlartiga Mar 19 '15 at 19:03
  • $\begingroup$ @rlartiga heh walked into that one still not the same as having max{0,xsinx) at least when I last TA'd we would expect the area bounded by the functions. Which would be abs(xsinx) $\endgroup$ – DRF Mar 19 '15 at 19:04
  • $\begingroup$ @DRF yes, if is asking the area bounded by the function should be $|x\sin{x}|$ $\endgroup$ – rlartiga Mar 19 '15 at 19:11

No it is not the same. First of all the area cannot be negative. Second of all, you have to divide it into two integrals. First one is the ,,positive" part where the graph of the function is above x axis and and the ,,negative" part where the graph is below x axis. The negative part has to be taken with absolute value. You have to count the integrals

$$Area=\int\limits_0^{\pi}x\sin xdx+|\int\limits_{\pi}^{2\pi}x\sin xdx| $$

Look at the graph of $x\sin x$.



No, the second part is a little different.

In the first question, you are evaluating the entire area of the function from x=0 to x=2pi, and you'll notice that you got a negative value.

This is because if you look at the graph of xsin(x), you'll see that at x=pi, the function goes from positive to negative. So in the first part you are adding up the entire area, (positive area)-(negative area).

In the second part you are given the constraint of the x axis, meaning they only want the positive area value between x=0 and x=2pi. Hope this helps you a little.

When given graphical constraints, it is helpful to graph the function and the lines you are given, and then visually understand what you are taking the integral of.


The function $xsin(x)$ is positive from $x=0$ to $x=\pi$ and negative from $x=\pi$ to $x=2\pi$. The value of definite integral is the algebraic sum of positive and negative areas in these two regions.
The area desired in second question could mean: $$\left|\int_{0}^{\pi}xsin(x)\right|+\left|\int_{\pi}^{2\pi}xsin(x)\right|$$


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