Integrals: Diagram-Based Questions (9)

hard 2 min read

Question

Find the area enclosed between the curves y=x2y = x^2 and y=2xx2y = 2x - x^2.

Solution — Step by Step

Set x2=2xx2x^2 = 2x - x^2:

2x22x=0    2x(x1)=0    x=0,12x^2 - 2x = 0 \implies 2x(x - 1) = 0 \implies x = 0, 1

So the curves meet at x=0x = 0 and x=1x = 1.

Pick a test point inside, say x=0.5x = 0.5. y1=0.25y_1 = 0.25, y2=10.25=0.75y_2 = 1 - 0.25 = 0.75. So y=2xx2y = 2x - x^2 lies above y=x2y = x^2 throughout [0,1][0, 1].

A=01[(2xx2)x2]dx=01(2x2x2)dxA = \int_0^1 \big[(2x - x^2) - x^2\big]\,dx = \int_0^1 (2x - 2x^2)\,dx A=[x22x33]01=123=13A = \left[x^2 - \frac{2x^3}{3}\right]_0^1 = 1 - \frac{2}{3} = \frac{1}{3}

Enclosed area =13= \dfrac{1}{3} square units.

Why This Works

The area between two curves is the integral of (top minus bottom) over the interval where they enclose a region. The intersections give the limits, and a quick test point tells us which is which.

If the curves cross multiple times within the interval, we must split the integral at each crossing and possibly take absolute values — but here they only meet at the endpoints.

Alternative Method

Geometric symmetry: the curve y=2xx2=1(x1)2y = 2x - x^2 = 1 - (x-1)^2 is a downward parabola with vertex at (1,1)(1, 1), and y=x2y = x^2 is an upward parabola through origin. The enclosed region is symmetric about x=1/2x = 1/2 — we can compute the integral from 00 to 1/21/2 and double it for a quick check.

Common Mistake

Forgetting to take “top minus bottom” — students sometimes integrate y1y2|y_1 - y_2| piecewise, which is fine, but if they integrate y2y1y_2 - y_1 thinking y1y_1 is on top, they get a negative answer and panic. Always identify top vs bottom with a test point first.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next