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

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Solution

$F r o m t h e h e l p o f f i g u r e A r e a o f B C F E = \int_{2}^{4} y d x W e k n o w t h a t y^{2} = 9 x ∴ y = \pm 3 \sqrt{x} S i n c e, I t i s i n t h e f i r s t q u a d r a n t ∴ y = 3 \sqrt{x} N o w A r e a o f B C F E = \int_{2}^{4} y d x = 3 \int_{2}^{4} \sqrt{x} d x = 3 \int_{2}^{4} {(x)}^{\frac{1}{2}} d x = 3 {⎡ ⎢ ⎢ ⎢ ⎣ \frac{x^{\frac{3}{2}}}{\frac{1}{2} + 1} ⎤ ⎥ ⎥ ⎥ ⎦}_{2}^{4} = 3 \times \frac{2}{3} {[x^{\frac{3}{2}}]}_{2}^{4} = 2 [{({(4)}^{\frac{1}{2}})}^{3} - {({(2)}^{\frac{1}{2}})}^{3}] = 2 [{(2)}^{3} - {(\sqrt{2})}^{3}] = 2 [8 - 2 \sqrt{2}] = 16 - 4 \sqrt{2} s q . u n i t s$

Hence, this is the answer.

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Find the area of the region bounded by y² = 9x, x = 2, x = 4 and the x-axis in the first quadrant.

y^{2} = 4 x, x = 1, and x = 4

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