How would I go about finding the area under a semicircle? I know that to use integration the formula is $\int_a^b f(x) \mathrm{d}x.$However, when I put this into my graphing calculator it doesn't work. If it matters, the points I am using for the semicircle are $(1,3.9)$ and $(3.5,3.9)$. The diameter is 2.5.
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$\begingroup$The idea here is to write a function for your equation and then integrate. I'll help you out with the function: We have a semicircle located $3.9$ units above the $x$ axis with center at $(2.25,3.9)$ and radius $1.25$. Hence the equation for the arc is
$$f(x)=3.9+\sqrt{1.25^2-(x-2.25)^2}$$
Now you just need to calculate the integral.
Here's a picture of the region:
The equation of a circle with radius $r$ is $x^2+y^2=r^2$. Solving for $y$ yields $y=\sqrt {r^2-x^2}$. This is a semicircle centered on the origin with radius $r$, to find the area of this semicircle, just integrate $y$ from one end of the semicircle to the other to have: $$\int_{-r}^{r}\sqrt {r^2-x^2} \,\mathrm{d}x=\frac {\pi r^2}{2}$$
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