Find the area of the region bounded by y^{2}= | vedclass

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(A) The area of the region bounded by the curve $y^{2}=9x$,the lines $x=2$ and $x=4$,and the $x$-axis in the first quadrant is given by the integral:A

Vedclass · Accepted Answer

$16-4\sqrt{2}$

Vedclass · Answer

(A) The area of the region bounded by the curve $y^{2}=9x$,the lines $x=2$ and $x=4$,and the $x$-axis in the first quadrant is given by the integral:Area $= \int_{2}^{4} y \, dx$Since $y^{2}=9x$ and we are in the first quadrant,$y = \sqrt{9x} = 3\sqrt{x}$.Area $= \int_{2}^{4} 3\sqrt{x} \, dx$$= 3 \int_{2}^{4} x^{1/2} \, dx$$= 3 \left[ \frac{x^{3/2}}{3/2} ight]_{2}^{4}$$= 3 	imes \frac{2}{3} \left[ x^{3/2} ight]_{2}^{4}$$= 2 \left[ 4^{3/2} - 2^{3/2} ight]$$= 2 \left[ 8 - 2\sqrt{2} ight]$$= 16 - 4\sqrt{2} 	ext{ square units}$.

Find the area of the region bounded by $y^{2}=9x$,$x=2$,$x=4$ and the $x$-axis in the first quadrant.

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