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The area in the first quadrant bounded by y=9x2y = 9x^2, x=0x = 0, y=1y = 1 and y=4y = 4 is :

Solution

✅ Correct Option: 4

Integrate with respect to yy since the region is bounded by y=1y=1 and y=4y=4. From y=9x2y=9x^2 in the first quadrant, x=y3x=\frac{\sqrt y}{3}. Area =∫14y3dy=29[y3/2]14=29(8−1)=149=\int_{1}^{4}\frac{\sqrt y}{3}dy=\frac{2}{9}\left[y^{3/2}\right]_1^4=\frac{2}{9}(8-1)=\frac{14}{9}.

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