Area between two curves

So far we have only found areas between the graph \(y=f(x)\) and the \(x\)-axis. In general we can find the area enclosed between two graphs \(y=f(x)\) and \(y=g(x)\). If \(f(x) > g(x)\), then the desired area is that which is below \(y=f(x)\) but above \(y=g(x)\), which is

\[ \int_a^b \big( f(x) - g(x) \big) \; dx. \]

This formula works regardless of whether the graphs are above or below the \(x\)-axis, as long as the graph of \(f(x)\) is above the graph of \(g(x)\). (Can you see why?)

Graphs of y = f(x) and y = g(x) shown with the graph of f above g in the first quadrant. A region between the two graphs is shaded.
Detailed description of diagram

In general, two graphs \(y=f(x)\) and \(y=g(x)\) may cross. Sometimes \(f(x)\) may be larger and sometimes \(g(x)\) may be larger. In order to find the area enclosed by the curves, we can find where \(f(x)\) is larger, and where \(g(x)\) is larger, and then take the appropriate integrals of \(f(x) - g(x)\) or \(g(x) - f(x)\) respectively.


Find the area enclosed between the graphs \(y=x^2\) and \(y=x\), between \(x=0\) and \(x=2\).

Graphs of y = x and y = x squared. Region shaded between the two graphs.
Detailed description of diagram


Solving \(x^2 = x\), we see that the two graphs intersect at \((0,0)\) and \((1,1)\). In the interval \([0,1]\), we have \(x \geq x^2\), and in the interval \([1,2]\), we have \(x^2 \geq x\). Therefore the desired area is

\begin{align*} \int_0^1 ( x - x^2 ) \; dx \, + \int_1^2 ( x^2 - x ) \; dx &= \Bigg[ \frac{1}{2} x^2 - \frac{1}{3} x^3 \Bigg]_0^1 + \Bigg[ \frac{1}{3} x^3 - \frac{1}{2} x^2 \Bigg]_1^2 \\ &= \Bigg( \Big( \frac{1}{2} - \frac{1}{3} \Big) - \big( 0 \big) \Bigg) + \Bigg( \Big( \frac{8}{3} - 2 \Big) - \Big( \frac{1}{3} - \frac{1}{2} \Big) \Bigg) \\ &= \Big( \frac{1}{6} - 0 \Big) + \Bigg( \frac{2}{3} - \Big( {-} \frac{1}{6} \Big) \Bigg) \\ &= \frac{1}{6} + \frac{5}{6} = 1. \end{align*}

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