The area bounded by a curve lying above the x-axis

2 The area bounded by a curve lying above the x-axis

Consider the graph of the function $y = f (x)$ shown in Figure 8. Suppose we are interested in calculating the area underneath the graph and above the $x$ -axis, between the points where $x = a$ and $x = b$ . When such an area lies entirely above the $x$ -axis, as is clearly the case here, this area is given by the definite integral $\int_{a}^{b} f (x) d x$ .

Figure 8

A graph showing the area bounded by a curve, the x-axis, and the vertical lines at x-coordinates a and b.

Long description

Key Point 4

The area under the curve $y = f (x)$ , between $x = a$ and $x = b$ is given by $\int_{a}^{b} f (x) d x$

when the curve lies entirely above the $x$ -axis between $a$ and $b$ .

Example 12

Calculate the area bounded $y = x^{- 1}$ and the $x$ -axis, between $x = 1$ and $x = 4$ .

Solution

Below is a graph of $y = x^{- 1}$ . The area required is shaded; it lies entirely above the $x$ -axis.

Figure 9

$A graph showing the area bounded by the curve $y = 1/x$, the x-axis, and the vertical lines $x=1$ and $x=5$.$

Long description

\begin{array}{rcl} area & = & \int_{1}^{4} \frac{1}{x} d x = {[ln |x|]}_{1}^{4} = ln 4 - ln 1 = ln 4 = 1.386 (3 d.p.) \end{array}

Task!

Find the area bounded by the curve $y = sin x$ and the $x$ -axis between $x = 0$ and $x = π$ . (The required area is shown in the figure. Note that it lies entirely above the $x$ -axis.)

$A graph showing the area bounded by the curve $y = \text{sin}(x)$ and the x-axis from $x=0$ to $x=\text{pi}$.$

Long description

$\int_{0}^{π} sin x d x = {[- cos x]}_{0}^{π} = 2$ .

Task!

Find the area under $f (x) = e^{2 x}$ from $x = 1$ to $x = 3$ given that the exponential function $e^{2 x}$ is always positive.

$area = \int_{1}^{3} e^{2 x} d x = {[\frac{1}{2} e^{2 x}]}_{1}^{3} = 198$ to 3 significant figures.

Example 13

The figure shows the graphs of $y = sin x$ and $y = cos x$ for $0 \leq x \leq \frac{1}{2} π$ . The two graphs intersect at the point where $x = \frac{1}{4} π$ . Find the shaded area.

Figure 10

$A graph showing the area bounded between the cosine function, $y = \text{cos}(x)$, and the sine function, $y = \text{sin}(x)$, from $x = -\frac{\pi}{2}$ to $x = \frac{\pi}{2}$.$

Long description

Solution

To find the shaded area we could calculate the area under the graph of $y = sin x$ for $x$ between 0 and $\frac{1}{4} π$ , and subtract this from the area under the graph of $y = cos x$ between the same limits. Alternatively the two processes can be combined into one and we can write

\begin{array}{rcl} shaded area & = & \int_{0}^{π ∕ 4} (cos x - sin x) d x \\ = & {[sin x + cos x]}_{0}^{π ∕ 4} \\ = & (sin \frac{1}{4} π + cos \frac{1}{4} π) - (sin 0 + cos 0) \\ = & (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1 \end{array}

So the numeric value of the integral is $\frac{2}{\sqrt{2}} - 1 = 0.414$ to 3 d.p.. (Alternatively you can use your calculator to obtain this result directly by evaluating $sin \frac{π}{4}$ and $cos \frac{π}{4}$ .)

Exercises

In each question you should check that the required area lies entirely above the horizontal axis.

Find the area under the curve $y = 7 x^{2}$ and above the $x$ -axis between $x = 2$ and $x = 5$ .
Find the area bounded by the curve $y = x^{3}$ and the $x$ -axis between $x = 0$ and $x = 2$ .
Find the area bounded by the curve $y = 3 t^{2}$ and the $t$ -axis between $t = - 3$ and $t = 3$ .
Find the area under $y = x^{- 2}$ between $x = 1$ and $x = 10$ .

273,
4,
54,
0.9.

2 The area bounded by a curve lying above the x-axis

Answer

Answer

Answer