Area of the region bounded by curve y^2=x,X-a | vedclass

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(A) The area $A$ of the region bounded by the curve $y^2=x$,the $X$-axis,and the lines $x=1$ and $x=4$ in the first quadrant is given by the definite

Vedclass · Accepted Answer

$\frac{14}{3}$

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(A) The area $A$ of the region bounded by the curve $y^2=x$,the $X$-axis,and the lines $x=1$ and $x=4$ in the first quadrant is given by the definite integral:$A = \int_{1}^{4} y \, dx$Since $y^2=x$ and we are in the first quadrant,$y = \sqrt{x} = x^{1/2}$.$A = \int_{1}^{4} x^{1/2} \, dx$Using the power rule for integration $\int x^n \, dx = \frac{x^{n+1}}{n+1}$:$A = \left[ \frac{x^{3/2}}{3/2} \right]_{1}^{4} = \left[ \frac{2}{3} x^{3/2} \right]_{1}^{4}$$A = \frac{2}{3} (4^{3/2} - 1^{3/2})$$A = \frac{2}{3} (8 - 1) = \frac{2}{3} (7) = \frac{14}{3}$ sq. units.Thus,the correct option is $A$.

Area of the region bounded by curve $y^2=x$,$X$-axis and lines $x=1$ and $x=4$ in the first quadrant is . . . . . . sq. units.

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