
Kenneth S. answered 02/10/17
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Calculus will seem easy if you have the right tutor!
The first part of the integrand is y=x and so the area under that line, on interval [0,1] is ½.
The second part of the integrand is y = √(1-x2) or x2+y2 = 1, a unit circle; now...what's the area under the circle in the first quadrant? Get it?