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

Question

Christopher Agresta · Accepted Answer

Area under curve is #5.57# sq.unit.

#y=f(x)= sqrt x#, area of the region bounded by the graph of #f(x)#, the x-axis and the vertical lines between #x=2 and x=5# is Area #A= int_a^b f(x)dx =int_2^5 x^(1/2) dx # or #A= [ x^(1/2+1)/(1/2+1)]_2^5 = 2/3 [x^(3/2)]_2^5# or #A=2/3 [5^(3/2)-2^(3/2)]=2/3(5 sqrt 5 - 2 sqrt 2)# or #A~~ 5.57# sq.unit . Area under curve is #5.57(2 d p)# sq.unit graph{x^(1/2) [-10, 10, -5, 5]} [Ans]

William Abdalla · Answer

undefined To find the area between the curve $ y = \sqrt{x} $ and the x-axis over the interval $[2,5]$, you integrate the absolute value of the function from 2 to 5:

$$ \text{Area} = \int_{2}^{5} |\sqrt{x}| \, dx $$

$$ \text{Area} = \int_{2}^{5} \sqrt{x} \, dx $$

$$ \text{Area} = \left[ \frac{2}{3}x^{3/2} \right]_{2}^{5} $$

$$ \text{Area} = \frac{2}{3}(5^{3/2} - 2^{3/2}) $$

$$ \text{Area} = \frac{2}{3}(125 - 8) $$

$$ \text{Area} = \frac{2}{3}(117) $$

$$ \text{Area} = 78 $$

Therefore, the area between the curve $ y = \sqrt{x} $ and the x-axis over the interval $[2,5]$ is $ 78 $ square units.