Area Between Two Curves

When two curves $y = f(x)$ and $y = g(x)$ enclose a region, the area between them is the integral of the difference (top minus bottom). It's just like finding the area under one curve, but vertically slicing the difference.

The Setup

If $f(x) \ge g(x)$ on $[a, b]$, the area is:

$A = \int_{a}^{b} \bigl(f(x) - g(x)\bigr)\,dx$

For our problem, the two curves are $y = 2x$ (the line) and $y = x^{2}$ (the parabola). Where do they intersect, and which is on top?

Step-by-Step Solution

Given: $f(x) = 2x$ and $g(x) = x^{2}$ on $[0, 2]$.

Find: The enclosed area.

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Step 1 — Find the intersection points.

Set $f(x) = g(x)$:

$2x = x^{2}$

$x^{2} - 2x = 0$

$x(x - 2) = 0$

$x = 0 \quad\text{or}\quad x = 2$

So the curves meet at $(0, 0)$ and $(2, 4)$ — exactly the endpoints of our interval.

Step 2 — Determine which curve is on top.

Pick a test point inside the interval, say $x = 1$:

$f(1) = 2(1) = 2 \qquad g(1) = 1^{2} = 1$

So $f(x) > g(x)$ on $(0, 2)$. The line is on top, the parabola is on bottom.

Step 3 — Set up the integral.

$A = \int_{0}^{2} \bigl(2x - x^{2}\bigr)\,dx$

Step 4 — Find the antiderivative.

$\int (2x - x^{2})\,dx = x^{2} - \dfrac{x^{3}}{3} + C$

Step 5 — Apply the Fundamental Theorem of Calculus.

$A = \left[x^{2} - \dfrac{x^{3}}{3}\right]_{0}^{2}$

$= \left((2)^{2} - \dfrac{(2)^{3}}{3}\right) - \left(0 - 0\right)$

$= 4 - \dfrac{8}{3}$

Step 6 — Simplify.

$A = \dfrac{12}{3} - \dfrac{8}{3} = \dfrac{4}{3}$

Step 7 — Decimal value.

$A = \dfrac{4}{3} \approx 1.333\;\text{square units}$

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Answer: The area enclosed between $y = 2x$ and $y = x^{2}$ on $[0, 2]$ is

$\boxed{\;A = \dfrac{4}{3} \approx 1.333\;}$

The line is above the parabola throughout the open interval $(0, 2)$, and the two curves meet at the endpoints. The shaded region is a "lens" between the line and the parabola.

Try It

• The shaded region is the answer — it sits between the cyan parabola and the pink line.
• Slide the left/right bounds to see how the area changes when you integrate over a sub-interval.
• The HUD shows the integrand $2x - x^{2}$ being summed numerically using a fine partition.