Answers>Maths>IB>Article

What is the area enclosed by the functions x^2 and sqrt(x)?

First, let's see how the plot of the functions looks like (draw on whiteboard). Next, let's calculate where the functions intersect by setting x2 = sqrt(x) and solving for x (manipulate by squaring both sides and get x4=x and combine to form x(x3-1)=0 which gives x=0 or 1). Finally, find the area by integrating the difference of the functions between these two points (integral from 0 to 1 of sqrt(x)-x2 dx = [2/3 x3/2 -1/3 x3] evaluated from 0 to 1 = 2/3-1/3 = 1/3). Therefore, the area enclosed by the functions x^2 and sqrt(x) is 1/3.

Answered by Maths tutor

1400 Views

See similar Maths IB tutors

Related Maths IB answers

All answers ▸

Prove that (sinx)^2 + (cosx)^2 = 1


f(x)=(2x+1)^0.5 for x >-0.5. Find f(12) and f'(12)


Let (x + 3) be a factor of the polynomial P(x) = x^3 + ax^2 - 7x + 6. Find a and the other two factors.


Find the coordinates of the minimum or maximum of the function f(x) = 3x^2 -2x +9 and determine if it's a minimum or maximum.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning