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The area of the region lying in the first quadrant and bounded by y=8x2,x=0,y=1y=8x^2, x=0, y=1 and y=4y=4 is:

Solution

✅ Correct Option: 1

Integrate with respect to yy using x=y/8=y22x=\sqrt{y/8}=\frac{\sqrt y}{2\sqrt2}. Area =∫14y22dy=122⋅23[y3/2]14=122⋅143=732=\int_1^4\frac{\sqrt y}{2\sqrt2}dy=\frac{1}{2\sqrt2}\cdot\frac{2}{3}\left[y^{3/2}\right]_1^4=\frac{1}{2\sqrt2}\cdot\frac{14}{3}=\frac{7}{3\sqrt2}.

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