How do you find the area between the given curve y= x^(1/2) and the x-axis given in the interval [2,5]?

Jun 9, 2018

Area under curve is $5.57$ sq.unit.

Explanation:

$y = f \left(x\right) = \sqrt{x}$, area of the region bounded by the graph of

$f \left(x\right)$, the x-axis and the vertical lines between $x = 2 \mathmr{and} x = 5$

is Area $A = {\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\int}_{2}^{5} {x}^{\frac{1}{2}} \mathrm{dx}$ or

$A = {\left[{x}^{\frac{1}{2} + 1} / \left(\frac{1}{2} + 1\right)\right]}_{2}^{5} = \frac{2}{3} {\left[{x}^{\frac{3}{2}}\right]}_{2}^{5}$ or

$A = \frac{2}{3} \left[{5}^{\frac{3}{2}} - {2}^{\frac{3}{2}}\right] = \frac{2}{3} \left(5 \sqrt{5} - 2 \sqrt{2}\right)$ or

$A \approx 5.57$ sq.unit .

Area under curve is $5.57 \left(2 d p\right)$ sq.unit

graph{x^(1/2) [-10, 10, -5, 5]}

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