Solve the integral from 0 to R of sqrt(R^2-x^2) dx

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The question was shade the region of the integral from 0 to R of sqrt(R^2-x^2) dx then evaluate the integral.

I shaded the circle to the right of the y axis, but I don't know how to integrate this problem. If it was 1/sqrt(R^2-x^2)dx, it would be an Arcsin result, but I've not seen an integral like this before. HELP!!
 
What you have is a quarter circle in the first quadrant.

I like trig substitution. Let's use it, OK?.

Let \(\displaystyle x=Rsin({\theta}) \;\ and \;\ dx=Rcos({\theta})d{\theta}\)

Make the subsitutions in your integral:

\(\displaystyle \L\\\int_{0}^{R}\sqrt{R^{2}-x^{2}}dx\)

\(\displaystyle \L\\\int_{0}^{R}\sqrt{R^{2}-(Rsin({\theta}))^{2}}\cdot{Rcos({\theta})}d{\theta}\)

\(\displaystyle \L\\\int_{0}^{R}\sqrt{R^{2}(1-sin^{2}({\theta}))}Rcos({\theta})d{\theta}\)

Don't forget to change your limits of integration:

\(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}R^{2}cos^{2}({\theta})d{\theta}\)

Can you finish?. You should know what the answer is beforehand, just by knowing how to find the area of a circle.
 
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