The area lying in the first quadrant and bounded by the circle $x^{2}+y^{2}=4$ and the lines $x= 0$ and $x=1$ is given by $\frac{\pi}{3}+\frac{\sqrt{3}}{2}$ $\frac{\pi}{6}+\frac{\sqrt{3}}{4}$ $\frac{\pi}{3}-\frac{\sqrt{3}}{2}$ $\frac{\pi}{6}+\frac{\sqrt{3}}{2}$

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Apr 23, 2018
in Calculus
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