1.8.2 Area Under the Graph

In this topic we will learn how to:

• use definite integration to find the area of a region bounded by a curve and lines parallel to the axes, or between a curve and a line or between two curves

We can use integration to find the area of a region. If the region lies between a curve and a straight line, assuming that the curve is on top, we subtract the area under the line from the area under the curve. The same would apply if the area lies between two curves.

Let’s walk through some past paper examples to understand how this works.

1. The equation of a curve is $y = 2\sqrt{3x + 4} - x$. Find the exact area of the region bounded by the curve, the $x$-axis and the lines $x = 0$ and $x = 4$. (9709/11/M/J/21 number 11)

As shown by the diagram, to get the area of the shaded region, we must integrate the curve from $0$ to $4$,
$$\textmd{Area } = \int_{0}^{4}\ 2\sqrt{3x + 4} - x\ dx$$$$\textmd{Area } = \int_{0}^{4}\ 2(3x + 4)^{\frac{1}{2}} - x\ dx$$Let’s integrate the curve,
$$\int_{0}^{4}\ 2(3x + 4)^{\frac{1}{2}} - x\ dx$$$$\left[\frac{2(3x + 4)^{\frac{3}{2}}}{3\left(\frac{3}{2}\right)} - \frac{x^{2}}{2}\right]_{0}^{4}$$$$\left[\frac{4}{9}(3x + 4)^{\frac{3}{2}} - \frac{x^{2}}{2}\right]_{0}^{4}$$Substitute in the limits,
$$\left[\frac{4}{9}(3x + 4)^{\frac{3}{2}} - \frac{x^{2}}{2}\right]_{0}^{4}$$$$\left[\left(\frac{4}{9}(3(4) + 4)^{\frac{3}{2}} - \frac{(4)^{2}}{2}\right) - \left(\frac{4}{9}(3(0) + 4)^{\frac{3}{2}} - \frac{(0)^{2}}{2}\right)\right]$$$$\left[\frac{184}{9} - \frac{32}{9}\right]$$$$\frac{152}{9}$$Therefore, the final answer is,
$$\textmd{Area } = \frac{152}{9}$$2. The diagram shows the curve with equation $y = (3x - 2)^{\frac{1}{2}}$ and the line $y = \frac{1}{2}x + 1$. The curve and the line intersect at the points $A(2, 2)$ and $B(6, 4)$. Find the area of the region enclosed by the curve and the line. (9709/11/M/J/22 number 7)

$$y = (3x - 2)^{\frac{1}{2}} \ \ \ \ y = \frac{1}{2}x + 1$$The curve is above the line, so we will subtract the area of the line from the area of the curve,
$$\int_{2}^{6} (3x - 2)^{\frac{1}{2}}\ dx - \int_{2}^{6} \frac{1}{2}x + 1\ dx$$To make the calculation less tedious, since the limits are the same, this can be written as,
$$\int_{2}^{6} (3x - 2)^{\frac{1}{2}} - \left(\frac{1}{2}x + 1\right)\ dx$$$$\int_{2}^{6} (3x - 2)^{\frac{1}{2}} - \frac{1}{2}x - 1\ dx$$Now let’s integrate,
$$\int_{2}^{6} (3x - 2)^{\frac{1}{2}} - \frac{1}{2}x - 1\ dx$$$$\left[\frac{(3x - 2)^{\frac{3}{2}}}{3\left(\frac{3}{2}\right)} - \frac{1}{2(2)}x^{2} - x\right]_{2}^{6}$$ $$\left[\frac{2}{9}(3x - 2)^{\frac{3}{2}} - \frac{1}{4}x^{2} - x\right]_{2}^{6}$$Substitute in the limits,

$$\left[\left(\frac{2}{9}(3(6) - 2)^{\frac{3}{2}} - \frac{1}{4}(6)^{2} - (6)\right) - \left(\frac{2}{9}(3(2) - 2)^{\frac{3}{2}} - \frac{1}{4}(2)^{2} - (2)\right)\right]$$

$$\left[-\frac{7}{9} - \left(-\frac{11}{9}\right)\right]$$$$\left[-\frac{7}{9} + \frac{11}{9}\right]$$$$\frac{4}{9}$$Therefore, the final answer is,
$$\textmd{Area } = \frac{4}{9}$$3. The diagram shows the curves with equations $y = x^{-\frac{1}{2}}$ and $y = \frac{5}{2} - x^{\frac{1}{2}}$. The curves intersect at the points $A\left(\frac{1}{4}, 2\right)$ and $B\left(4, \frac{1}{2}\right)$. Find the area of the region between the two curves. (9709/13/O/N/21 number 8)

$$y = x^{-\frac{1}{2}} \ \ \ \ y = \frac{5}{2} x^{\frac{1}{2}}$$The curve $y = \frac{5}{2} - x^{\frac{1}{2}}$ is above the curve $y = x^{-\frac{1}{2}}$, so subtract the area under $y = x^{-\frac{1}{2}}$ from the area under $y = \frac{5}{2 -} x^{\frac{1}{2}}$,
$$\int_{\frac{1}{4}}^{4}\ \left(\frac{5}{2} - x^{\frac{1}{2}}\right) - x^{-\frac{1}{2}}\ dx$$$$\int_{\frac{1}{4}}^{4}\ \frac{5}{2} - x^{\frac{1}{2}} - x^{-\frac{1}{2}}\ dx$$Now let’s integrate,
$$\int_{\frac{1}{4}}^{4}\ \frac{5}{2} - x^{\frac{1}{2}} - x^{-\frac{1}{2}}\ dx$$$$\left[\frac{5}{2}x - \frac{2}{3}x^{\frac{3}{2}} - 2x^{\frac{1}{2}}\right]_{\frac{1}{4}}^{4}$$Substitute in the limits,

$$\left[\left(\frac{5}{2}\left(4\right) - \frac{2}{3}\left(4\right)^{\frac{3}{2}} - 2\left(4\right)^{\frac{1}{2}}\right) - \left(\frac{5}{2}\left(\frac{1}{4}\right) - \frac{2}{3}\left(\frac{1}{4}\right)^{\frac{3}{2}} - 2\left(\frac{1}{4}\right)^{\frac{1}{2}}\right)\right]$$

$$\left[\frac{2}{3} - \left(-\frac{11}{24}\right)\right]$$$$\left[\frac{2}{3} + \frac{11}{24}\right]$$$$\frac{9}{8}$$Therefore, the final answer is,
$$\textmd{Area } = \frac{9}{8}$$