Integrals: Application Problems (5)

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Question

Find the area enclosed between the parabola y=x2y = x^2 and the line y=x+2y = x + 2.

Solution — Step by Step

Set x2=x+2x^2 = x + 2:

x2x2=0(x2)(x+1)=0x=1,2x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1) = 0 \Rightarrow x = -1, 2

So the curves meet at x=1x = -1 and x=2x = 2.

Pick a test point in between, say x=0x = 0. Line: y=2y = 2. Parabola: y=0y = 0. So the line is above the parabola throughout the interval [1,2][-1, 2].

A=12[(x+2)x2]dxA = \int_{-1}^{2} \left[(x + 2) - x^2\right] dx A=[x22+2xx33]12A = \left[\tfrac{x^2}{2} + 2x - \tfrac{x^3}{3}\right]_{-1}^{2}

At x=2x = 2: 2+483=683=1032 + 4 - \tfrac{8}{3} = 6 - \tfrac{8}{3} = \tfrac{10}{3}.

At x=1x = -1: 122+13=312+26=76\tfrac{1}{2} - 2 + \tfrac{1}{3} = \tfrac{3 - 12 + 2}{6} = -\tfrac{7}{6}.

A=103(76)=206+76=276=92A = \tfrac{10}{3} - \left(-\tfrac{7}{6}\right) = \tfrac{20}{6} + \tfrac{7}{6} = \tfrac{27}{6} = \tfrac{9}{2}

Area = 9/2=4.59/2 = 4.5 square units.

Why This Works

To find area between two curves, integrate (upper − lower) with respect to xx. The bounds are the xx-coordinates of intersection. If the upper-lower relation flips, split the integral at the crossover point and integrate each piece separately.

The intuition: at each xx in the interval, the vertical strip has height (upper − lower) and width dxdx, so its area is the integrand. Adding all strips via integration gives the total enclosed area.

Alternative Method

Integrate with respect to yy instead — invert both curves: x=±yx = \pm\sqrt{y} for the parabola, x=y2x = y - 2 for the line. Set up two horizontal-strip integrals. More work in this case, useful when the curves are easier to express in x=f(y)x = f(y) form.

Common Mistake

Students forget to verify which curve is on top. If you integrate (parabola − line) instead of (line − parabola), you get 9/2-9/2 — area can never be negative. Either check at a test point first, or take the absolute value at the end. Also: never just compute 12x2(x+2)dx\int_{-1}^2 |x^2 - (x+2)| dx blindly without knowing the order — the absolute value handles sign but you still need the right order to set it up.

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