Definite integrals, signed area, and area
Up to this point, we have only considered the calculation of definite integrals in regions where all function values are positive and this represented the area under the under the curve over the given interval. What do definite integrals represent when the function is partly or fully below the horizontal axis?
Consider the function $f(x)=x^2+2x-3$f(x)=x2+2x−3 between $x=-3$x=−3and $x=1$x=1:
We can see this area is completely below the horizontal axis. Setting up the integral we get:
We can see that the value of the definite integral is negative. As the result of the definite integral has a sign depending on whether the area is above or below the $x$x-axis, we call this the signed area.
In this case, the actual area bounded by the curve and the $x$x-axis is equivalent to the absolute value of the integral. That is, the area is $\frac{32}{3}$323 units2.
We can demonstrate the last property of definite integrals using areas under a curve:
Split interval:
This property is very useful to combine to integrals with a shared start/end point or to split a given integral into separate regions.
We can split an integral over a closed interval $\left[a,c\right]$[a,c] at the point $b$b where $b$b is within the closed interval, the sum of the integral of the parts equals the original integral.
This can be shown from the definition of a definite integral as follows:
For areas under a curve we can also visualise this property as:
We can see in this case, where $a\le b\le c$a≤b≤c and $f\left(x\right)\ge0$f(x)≥0, we have:
When evaluating a definite integral over a given interval, it may include sections that are above and below the horizontal axis, which have signed area that is positive and negative respectively. Evaluating the definite integral in one go gives the overall signed area - that is, the sum of all the positive and negative regions (signed areas) over the interval of integration. This means that we could get a positive value, a negative value, or even zero when evaluating the definite integral.
However, if we want to calculate the area under a curve over an interval, this is different from the signed area. In order to calculate the actual area between the curve and the $x$x-axis, we can divide the interval into sections that are fully above and sections that are fully below the horizontal axis, and then calculate these integrals individually. We then take the absolute value of each of the regions (that is, make all of the negative ones positive) and add them together to find the total area.
The example below highlights the difference between the calculation of signed area and area.
Worked example
Example 1
For the function $y=(x+2)^3-1$y=(x+2)3−1:
(a) Evaluate the integral of the function between $x=-3$x=−3 and $x=0$x=0.
Think: We are asked to evaluate a definite integral over a given integral, hence, we are calculating the signed area.
Do: We evaluate the following integral between $x=-3$x=−3 and $x=0$x=0:
(b) Calculate the area bounded by the function and the $x$x axis between $x=-3$x=−3 and $x=0$x=0
Think: we want to calculate the area over a given integral. Hence, we need to take care with positive and negative regions.
Do: Sketching the function over the given interval will tell us if there are going to be in any negative regions.
Using the calculator or the given form of the equation - the graph could be obtained from $y=x^3$y=x3 by a horizontal translation to the left by $2$2 units and then a vertical translation of $1$1 unit downwards, we obtain the following graph:
The area between $x=-3$x=−3 and $x=-1$x=−1 is below the horizontal axis and between $x=-1$x=−1 and $x=0$x=0 is above. So we will split our integral into two sections and take the negative of the integral for the part under the axis, as follows:
Hence the area is $4\frac{3}{4}$434 units2.
Reflect: Notice that the value for the area calculated in part (b) is greater than the signed area. This is because the signed area is the sum of positive and negative regions, whereas calculating the area takes the absolute value of each region and sums them together. Note that the integral for the negative region was given a negative sign to make it positive. Taking the absolute value would have achieved the same result.
Use the following applet to explore the overall signed area of an odd function.
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Practice questions
Question 1
Consider the function $f\left(x\right)$f(x) as shown. The numbers in the shaded regions indicate the area of the region.
Determine $\int_{-3}^0f\left(x\right)dx$∫0−3f(x)dx.
Determine the area enclosed by the curve and the $x$x-axis for $x<0$x<0.
Determine $\int_{-3}^2f\left(x\right)dx$∫2−3f(x)dx.
Determine the area enclosed by the curve and the $x$x-axis.
Question 2
Consider the function $y=x\left(x-1\right)\left(x+3\right)$y=x(x−1)(x+3).
Graph the function.
Loading Graph...Hence determine the exact area bounded by the curve and the $x$x-axis.
Question 3
Consider the function $f\left(x\right)$f(x) as shown. The numbers in the shaded regions indicate the area of the region.
Write the following expressions in terms of $A$A and $B$B.
$\int_{-5}^1f\left(x\right)dx$∫1−5f(x)dx
$\int_{-5}^{-2}3f\left(x\right)dx-\int_{-2}^1f\left(x\right)dx$∫−2−53f(x)dx−∫1−2f(x)dx
$\left|\int_{-5}^1f\left(x\right)dx\right|$|∫1−5f(x)dx|
$\int_{-5}^1\left|f\left(x\right)\right|dx$∫1−5|f(x)|dx
$\int_{-5}^1\left(f\left(x\right)+x\right)dx$∫1−5(f(x)+x)dx given that $\int_{-5}^1xdx=-12$∫1−5xdx=−12