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

8937 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-09T12:43:12

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]