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My book "explains" how to find the area of a circle using integration by first finding the area of a semicircle:

take a semicircle with the radius $r$ , such as the one below:

graph

$f(x)= \sqrt{r^2-x^2}$ is the function of this semi-circle, and to find its area:

$A= \int _{-r}^{r} \sqrt{r^2-x^2}$

we substitute $x=r*cos(t)$

between $t=0 \Rightarrow x=r*cos(0)=r $

and $t= \pi \Rightarrow x=r*cos(pi)= -r$ .

and $dx = -r*sin(t)dt$

and rewrite the integral as:

$A = \int _{\pi}^0 \sqrt{r^2-r^2*cos^2(t)} (-r*sin(t))dt$

$A = \int _{\pi}^0 -r^2* \sqrt{1-cos^2(t)}sin(t)dt$

$A = r^2 \int _{0}^{\pi} sin^2(t)dt = \frac{\pi r^2}{2}$

where did the negative sign of $-r^2$ go?? and how did the limits get inversed (ie $0$ becoming the lower limit and $\pi$ becoming the upper limit) ??

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    $\begingroup$ Switching upper and lower limits changes the sign, so your two questions answer each other by cancelling each other out. $\endgroup$ – Ethan Bolker Aug 31 '17 at 12:59
  • $\begingroup$ really? first time I've heard of that but that makes it make sense, thanks! $\endgroup$ – Dahen Aug 31 '17 at 12:59
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The following might explain the matter $$\int_a^b f(x) \mathrm{d}x = - \int_b^a f(x) \mathrm{d}x$$

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  • $\begingroup$ yeah it seems kind of obvious now that I look at it tbh lol, thanks for the help! $\endgroup$ – Dahen Aug 31 '17 at 13:00

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