Areas Between Curves Examples 1

Recall that for two continuous curves $y = f(x)$ and $y = g(x)$, if f(x) ≥ g(x) for a ≤ x ≤ b, then it follows that the area trapped between these two curves is:

(1)
\begin{align} A = \int_a^b [ f(x) - g(x) ] \: dx \end{align}

Similarly, for two continuous curves $x = f(y)$ and $x = g(y)$, if f(y) ≥ g(y) for c ≤ d, then it follows that the area trapped between these two curves is:

(2)
\begin{align} A = \int_c^d [ f(y) - g(y) ] \: dy \end{align}

We will now apply this technique in the following examples.

## Example 1

Determine the area trapped between the curves $f(x) = 5x - x^2$ and $g(x) = x$.

We note that these two graphs intersect each other at x = 0 and x = 4. Hence our limits of integration will be 0 and 4. Furthermore, we note that f(x) ≥ g(x) for 0 ≤ x ≤ 4 as shown when graphed below: We now want to determine the area of the shaded region, which we can do by using the formula at the top of the page:

(3)
\begin{align} A = \int_0^4 [(5x - x^2) - (x)] \: dx \\ A = \int_0^4 [ 4x - x^2 ] \: dx \\ A = ( 2x^2 - \frac{x^3}{3} ) \bigg |_0^4 \\ A = 32 - \frac{64}{3} \\ A = \frac{96}{3} - \frac{64}{3} \\ A = \frac{32}{3} \end{align}

## Example 2

Calculate the area trapped between the curves $f(x) = x$, $g(x) = x^2$.

We first note that f(x) and g(x) intersect at x = 0 and x = 1. Hence our limits of integration are 0 and 1. Furthermore, we note that f(x) ≥ g(x) for 0 ≤ x ≤ 1 as shown when the graphs are drawn: Hence it follows that:

(4)
\begin{align} A = \int_0^1 [(x) - (x^2)] \: dx \\ A = \left ( \frac{x^2}{2} - \frac{x^3}{3} \right ) \bigg |_0^1 \\ A = \frac{1}{2} - \frac{1}{3} \\ A = \frac{1}{6} \end{align}

## Example 3

Calculate the area trapped between the curves $f(x) = 1/x$, $g(x) = 1/x^2$, and $x = 2$.

We note that the curves f(x) and g(x) intersect at x = 1. We were also given the upper limit of integration. Hence our limits of integration are 1 and 2. Furthermore, we note that f(x) ≥ g(x) for 1 ≤ x ≤ 2 (which we can verify by testing any value of x on this interval). The area we are trying to calculate is shown in the following graph: All we need to do is evaluate the following integral:

(5)
\begin{align} A = \int_1^2 [ (x^{-1}) - (x^{-2}) ] \: dx \\ A = (\ln x + x^{-1}) \bigg |_1^2 \\ A = \left( \ln (2) + \frac{1}{2} \right ) - (\ln (2) + 1 )\\ A = \ln (2) - \frac{1}{2} \end{align}

## Example 4

Calculate the are trapped between the curves $f(y) = 2y^2$ and $g(y) = 4 + y^2$.

We first note that these two curves intersect when y = -2 and y = 2. Hence our limits of integration are -2 and 2. Furthermore, we note that g(y) ≥ f(y) for -2 ≤ y ≤ 2. Hence we can evaluate this integral as follows:

(6)
\begin{align} A = \int_{-2}^2 [(4 + y^2) - (2y^2)] \: dy \\ A = \int_{-2}^2 [ 4 - y^2 ] \: dy \\ A = \left ( 4y - \frac{y^3}{3} \right ) \bigg |_{-2}^{2} \\ A = \left ( 8 - \frac{8}{3} \right ) - \left ( -8 + \frac{8}{3} \right ) \\ A = 16 - \frac{16}{3} \\ A = \frac{32}{3} \end{align}