Title: Default for 13.3 The Area Bounded by a Curve
Course: HELM Workbooks 2023-2024

The area bounded by a curve

One of the important applications of integration is to find the area bounded by a curve. Often such an area can have a physical significance like the work done by a motor, or the distance travelled by a vehicle. In this Section we explain how such an area is calculated.

Exercise

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

1. Find the area under the curve \(y=[value]\) and above the x-axis between \(x = [value]\) and \(x = [value]\).
2. Find the area bounded by the curve \(y=[value]\) and the x-axis between \(x = [value]\) and \(x = [value]\).
Feedback: Since the required area lies entirely above the horizontal axis for each question, the area under the curve \(y\) between \(x = a\) and \(x = b\) is \(\displaystyle\int_{a}^{b} y(x) \, \mathrm{d}x\). 1. \(\displaystyle \int_[value]^[value][value]\, \mathrm{d}x = \left[ [value] \right]_[value]^[value] = [value]\). A sketch of the graph is shown below. 2. \(\displaystyle \int_[value]^[value][value]\, \mathrm{d}x =\left[ [value] \right]_[value]^[value] = [value]\). A sketch of the graph is shown below.

Calculating the area under a curve

Let us denote the area under \(y=f(x)\) between a fixed point \(a\) and a variable point \(x\) by \(A(x)\):

[Image: Area under a generic function]

\(A(x)\) is clearly a function of \(x\) since as the upper limit changes so does the area. How does the area change if we change the upper limit by a very small amount \(\delta x\)? See the figure below.

[Image: Area under a generic function ii]

To a good approximation the change in the area is: \(A(x+\delta x)-A(x)\approx f(x)\delta x\) [This is because the shaded area is approximately a rectangle with base \(\delta x\) and height \(f(x)\).]
This approximation gets better and better as \(\delta x\) gets smaller and smaller.
Rearranging gives: \(f(x)\approx \dfrac{A(x+\delta x)-A(x)}{\delta x}\)
Clearly, in the limit as \(\delta x\to 0\) we have \(f(x)={\displaystyle\lim_{\delta x\to 0}} \dfrac{A(x+\delta x)-A(x)}{\delta x}\)
But this limit on the right-hand side is the derivative of \(A(x)\) with respect to \(x\) so \(f(x)=\dfrac{d A(x)}{d x}\)
Thus \(A(x)\) is an indefinite integral of \(f(x)\) and we can therefore write: \(A(x)={\displaystyle\int f(x)\ \mathrm{d}x}\)
Now the area under the curve from \(a\) to \(b\) is clearly \(A(b)-A(a)\).

But remembering our shorthand notation for this difference, introduced in the last Section we have, finally \(A(b)-A(a)\equiv\left[\phantom{\dfrac{1}{1}}\!\!\!A(x)\right]_a^b={\displaystyle\int_a^bf(x)\ \mathrm{d} x}\)
We conclude that the area under the curve \(y=f(x)\) from \(a\) to \(b\) is given by the definite integral of \(f(x)\) from \(a\) to \(b\).

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

Consider the graph of the function \(y=f(x)\) shown in the figure below. 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 \({\displaystyle\int_a^b f(x)\ \mathrm{d}x}\).

[Image]

Key Point

The area under the curve \(y=f(x)\), between \(x=a\) and \(x=b\) is given by \({\displaystyle\int_a^b f(x)\ \mathrm{d}x}\) when the curve lies entirely above the \(x\) axis between \(a\) and \(b\) .

Example

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.

[Image]

\({\displaystyle\int_1^4 \dfrac{1}{x}\ \mathrm{d}x}=\bigg[ln |x|\bigg]_1^4= \ln 4 - \ln 1 = \ln 4= 1.386\) (\(3\) d.p.)

The area bounded by a curve, not entirely above the x-axis

The following figure shows a graph of \(y=-x^2+1\).

[Image: Area between x-axis and y = -x^2 + 1 between 1 and 2]

The shaded area is bounded by the \(x\)-axis and the curve, but lies entirely below the \(x\)-axis.</p>

Let us evaluate the integral \({\displaystyle\int_1^2( -x^2+1)\ \mathrm{d}x}\).

\[\begin{aligned} {\displaystyle\int_1^2( -x^2+1)\ \mathrm{d}x} = \left[ -\dfrac{x^3}{3}+x\right]_1^2 &= \left(-\dfrac{2^3}{3} + 2 \right) - \left(-\dfrac{1^3}{3} + 1 \right), \\ &= -\dfrac{7}{3} + 1, \\ &= -\dfrac{4}{3}.  \end{aligned}\]

The evaluation of the area yields a negative quantity. There is, of course, no such thing as a negative area. The area is actually \(\dfrac{4}{3}\), and the negative sign is an indication that the area lies below the \(x\)-axis. (However, in applications of integration such as work/energy or distance travelled in a given direction negative values can be meaningful.)

If an area contains parts both above and below the horizontal axis, care must be taken when calculating this area. It is necessary to determine which parts of the graph lie above the horizontal axis and which lie below. Separate integrals need to be calculated for each ‘piece’ of the graph. This idea is illustrated in the next Example.

Example

Find the total area enclosed by the curve \(y=x^3-5x^2+4x\) and the \(x\)-axis between \(x=0\) and \(x=3\).

Solution

We need to determine which parts of the graph lie above and which lie below the \(x\)-axis. To do this it is helpful to consider where the graph cuts the \(x\)-axis. So we consider the function \(x^3-5x^2+4x\) and look for its zeros

\[x^3-5x^2+4x = x(x^2-5x+4) = x(x-1)(x-4)\]

So the graph cuts the \(x\)-axis when \(x=0\), \(x=1\) and \(x=4\). Also, when \(x\) is large and positive, \(y\) is large and positive since the term involving \(x^3\) dominates. When \(x\) is large and negative, \(y\) is large and negative for the same reason. With this information we can sketch a graph showing the required area:

[Image: the area between the x-axis and the function y = x^3 -5x^2+1 between zero and four]

From the graph we see that the required area lies partly above the \(x\)-axis (when \(0\le x\le 1\)) and partly below (when \(1\le x\le 3\)). So we evaluate the integral in two parts:

Firstly:

\[{\displaystyle\int_0^1( x^3-5x^2+4x )\ \mathrm{d}x} =  \left[\dfrac{x^4}{4} - \dfrac{5x^3}{3} + \dfrac{4x^2}{2}\right]_0^1 = \left(\dfrac{1}{4} - \dfrac{5}{3} + 2\right) - \left(0\right) = \dfrac{7}{12}\]

This is the part of the required area which lies above the \(x\)-axis.

Secondly:

\[\begin{aligned} {\displaystyle\int_1^3( x^3-5x^2+4x )\ \mathrm{d}x} &= \left[\dfrac{x^4}{4} - \dfrac{5x^3}{3} + \dfrac{4x^2}{2}\right]_1^3, \\ &= \left(\dfrac{81}{4} - \dfrac{135}{3} + 18\right) - \left(\dfrac{1}{4} - \dfrac{5}{3} + 2\right), \\ &= -\dfrac{22}{3}. \end{aligned}\]

This represents the part of the required area which lies below the \(x\)-axis.

The actual area is \(\dfrac{22}{3}\).

Combining the results of the two separate calculations we can find the total area bounded by the curve:

area = \(\dfrac{7}{12} + \dfrac{22}{3} = \dfrac{95}{12}\)

Q: Exercise

1. Find the total area enclosed between the \(x\)-axis and the curve \(y=[value]\) between \(x=[value]\) and \(x=[value]\).
2. Fnd the area under \(y=[value]\) from \(t=[value]\) to \(t=[value]\). Give your answer to \(4\) significant figures.
3. Find the area enclosed by \(y=[value]\) and the \(x\)-axis from \(x=[value]\) to \(x=[value]\).
Feedback: 1. Since the curve \(y=[value]\) is above the \(x\) axis between 0 and [value], and below the \(x\) axis between [value] and 0. The area under the curve is: \[-\displaystyle\int_[value]^{0} [value] \, \mathrm{d}x + \displaystyle\int_{0}^[value] [value] \, \mathrm{d}x = [value].\] 2. The curve \(y=[value]\) is above the \(t\) axis from \(t=[value]\) to \(t=[value]\). The area is: \[\displaystyle\int_[value]^[value] [value] \, \mathrm{d}t = \left[ [value] \right]_[value]^[value] = [value].\] 3. The area under the curve \(y=[value]\) is above the \(x\) axis from \(x=[value]\) to \(x=[value]\), hence the area is: \[\displaystyle \int_[value]^[value] ([value]) \, \mathrm{d}x =\left[ [value] \right]_[value]^[value] = [value].\]
  Hint: If you are unsure of the limits of integration draw a diagram.

Exercise

Find the area under \(y = \displaystyle[value]\) between \(x = [value]\) and \(x = [value]\).

You should check that the required area lies entirely above the horizontal axis.
Feedback: Since the required area lies entirely above the horizontal axis for each question, the area under the curve \(y\) between \(x = a\) and \(x = b\) is \(\displaystyle\int_{a}^{b} y(x) \, \mathrm{d}x\). \(\displaystyle \int_[value]^[value] [value] \, \mathrm{d}x = \left[ [value] \right]_[value]^[value] = [value].\) A sketch of the graph is shown below.

Exercise

Find the area bounded by the curve \(y = [value]\) and the t-axis between \(t = [value]\) and \(t = [value]\).

You should check that the required area lies entirely above the horizontal axis.
Feedback: Since the required area lies entirely above the horizontal axis for each question, the area under the curve \(y\) between \(t = a\) and \(t = b\) is \(\displaystyle\int_{a}^{b} y(t) \, \mathrm{d}t\). \(\displaystyle \int_[value]^[value] [value] \, \mathrm{d}t = \left[ [value] \right]_[value]^[value] = [value].\) A sketch of the graph is shown below.

Q: Exercise

Calculate the area enclosed by the curve \(y=x^3\) and the line \(y=x\).
Feedback: The curve \(y=x^3\) and the line \(y=x\) intersect at \(x=-1,0,1\). Between 0 and 1 the graph is above the \(x\) axis and \(y=x\) is greater than \(y=x^3\). Between -1 and 0, the graph is below the \(x\) axis and \(y=x^3\) is greater than \(y=x\). Hence the area is \[- \int_{-1}^0 (x^3 -x) \,\mathrm{d}x + \int_0^1 (x-x^3) \, \mathrm{d}x = [value]\] A sketch of the graph is shown below.
  Hint: If you are unsure of the limits of integration draw a diagram.

Q: Exercise

1. Find the area bounded by \(y= e^x\), the \(y\)-axis and the line \(x=2\).
2. Find the area enclosed between \(y=x(x-[value])(x-[value])\) and the \(x\)-axis.
Feedback: 1. The curve \(y= e^x\), the \(y\)-axis and the line \(x=2\) intersect at \(x=0,2\).This area is above the \(x\) axis hence the area is \[\displaystyle\int_0^2 e^x \, \mathrm{d}x = [value].\] A sketch of the curve \(y = [value]\) between \(x=0\) an \(x=2\) is shown below. 2. The curve \(y=[value]\) and the \(x\) axis intersect at \(x = 0,[value],[value]\). Between \(0\) and [value], the area under the curve is above the \(x\) axis and between [value] and [value] , it is below the \(x\) axis. Hence the area is: \[\displaystyle\int_[value]^[value] x(x-[value])(x-[value]) \, \mathrm{d}x - \displaystyle\int_[value]^[value] x(x-[value])(x-[value]) \, \mathrm{d}x =\] \[\displaystyle\int_[value]^[value] ([value]) \, \mathrm{d}x - \displaystyle\int_[value]^[value]([value]) \, \mathrm{d}x =\] \[ \left[[value] \right]_[value]^[value] -\left[[value]\right]_[value]^[value] = [value] .\] A sketch of the graph is shown below. [value]
  Hint: If you are unsure of the limits of integration draw a diagram.

Exercise

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

[Image: Area under the sine of x between 0 and pi]

Area \(=\)
Feedback: \({\displaystyle\int_0^\pi \sin x\ \mathrm{d}x}=\bigg[-\cos x\bigg]_0^\pi = -\cos(\pi) - (- \cos (0)) = 2.\)

Exercise

Find the area under \(f(x)=[value]\) from \(x=[value]\) to \(x=[value]\) given that the exponential function \([value]\) is always positive. Give your answer in its exact form.

Area \(=\)
Feedback: The following graph shows the graph of the function \(f(x)=[value]\) and the area we want to find. Area =\({\displaystyle\int_1^3 [value]\ \mathrm{d}x}\) = \(\bigg[\displaystyle [value]\bigg]_1^3=[value]\).

Exercise

In the image you can see the sketch of the graph \( y= \sin 2x\) for \( 0 \leq x \leq \pi.\)

[Image]

Find the total area bounded by the curve and the \(x\)-axis between \(x= \dfrac{1}{3}\pi\) and \(x=\dfrac{3}{4}\pi\).

Area \(=\)
Feedback: [Image] Perform the integration in two parts to obtain the required area: \({\displaystyle\int_\frac{\pi}{3}^\frac{\pi}{2} \sin(2x)\ \mathrm{d}x} = \dfrac{1}{4}\) and \({\displaystyle\int_\frac{\pi}{2}^\frac{3\pi}{4} \sin(2x)\ \mathrm{d}x} = -\dfrac{1}{2}\). The required area is \(\dfrac{1}{4} + \dfrac{1}{2} = \dfrac{3}{4}\).

Example

The figure shows the graphs of \(y=\sin x\) and \(y=\cos x\) for \({0\le x\le \dfrac{1}{2}\pi}\). The two graphs intersect at the point where \(x=\dfrac{1}{4}\pi\). Find the shaded area.

[Image]

Solution

To find the shaded area we could calculate the area under the graph of \(y=\sin x\) for \(x\) between 0 and \(\dfrac{1}{4} \pi\), 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

shaded area \(\\={\displaystyle\int_0^{\pi/4}( \cos x - \sin x )\ \mathrm{d}x} \\= \bigg[ \sin x + \cos x\bigg]_0^{\pi/4} \\= \left(\sin \left(\tfrac{1}{4}\pi\right) + \cos \left(\tfrac{1}{4}\pi\right)\right) -\left( \sin 0 +\cos 0\right) \\= \left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\right)-(0+1)\\=\dfrac{2}{\sqrt{2}}-1\\=\sqrt{2}-1\)