Definite integrals, signed area and area
We saw in the previous lesson, functions above the horizontal axis will return positive values when integrated and functions below the axis will return negative values when integrated. The evaluation of a definite integral over a closed interval that includes both positive and negative function values is called the signed area. The signed area is the sum of all positive and negative values of the integral. This means that we can get negative values or even zero when evaluating integrals.
If we want to find the actual area between a function and an axis, we need to divide the interval into positive and negative regions and then calculate the integrals individually. We then either multiply negative regions by a negative to make them positive or take the absolute value of the regions 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:
- Evaluate the integral between $x=-3$x=−3 and $x=0$x=0.
- Calculate the area bounded by the function and the $x$x-axis 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:
| $\int_{-3}^0\ (x+2)^3-1\ dx$∫0−3 (x+2)3−1 dx | $=$= | $[\frac{(x+2)^4}{4}-x]_{-3}^0$[(x+2)44−x]0−3 |
| $=$= | $[(\frac{(0+2)^4}{4}-0-(\frac{(-3+2)^4}{4}-(-3))]$[((0+2)44−0−((−3+2)44−(−3))] | |
| $=$= | $[4-(\frac{1}{4}+3)]$[4−(14+3)] | |
| $=$= | $\frac{3}{4}$34 |
Think: We are asked 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 what we know about graphing cubic functions we get 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:
| $\int_{-3}^0\ (x+2)^3-1\ dx$∫0−3 (x+2)3−1 dx | $=$= | $-\int_{-3}^{-1}\ (x+2)^3-1\ dx+\int_{-1}^0\ (x+2)^3-1\ dx$−∫−1−3 (x+2)3−1 dx+∫0−1 (x+2)3−1 dx |
| $=$= | $-[\frac{(x+2)^4}{4}-x]_{-3}^{-1}+[\frac{(x+2)^4}{4}-x]_{-1}^0$−[(x+2)44−x]−1−3+[(x+2)44−x]0−1 | |
| $=$= | $-[\frac{(-1+2)^4}{4}-(-1)-(\frac{(-3+2)^4}{4}-(-3))]+[\frac{(0+2)^4}{4}-0-(\frac{(-1+2)^4}{4}-(-1))]$−[(−1+2)44−(−1)−((−3+2)44−(−3))]+[(0+2)44−0−((−1+2)44−(−1))] | |
| $=$= | $-[\frac{1}{4}+1-\frac{1}{4}-3)]+[4-(\frac{1}{4}+1)]$−[14+1−14−3)]+[4−(14+1)] | |
| $=$= | $2+2\frac{3}{4}$2+234 | |
| $=$= | $4\frac{3}{4}$434 |
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 negative region, in this case, was multiplied by a negative to make it positive. Taking the absolute value achieves the same result.
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 $f\left(x\right)$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, $B$B and $C$C.
$\int_0^6f\left(x\right)dx$∫60f(x)dx
$\int_6^{-1}f\left(x\right)dx$∫−16f(x)dx
The area bounded by the curve $f\left(x\right)$f(x) and the $x$x-axis
$\int_{-1}^6\left|f\left(x\right)\right|dx$∫6−1|f(x)|dx
$\int_{-1}^6\left(2x-f\left(x\right)\right)dx$∫6−1(2x−f(x))dx given that $\int_{-1}^62xdx=35$∫6−12xdx=35
Properties of definite integrals
Definite integrals have a number of properties that can assist in simplifying and solving problems involving area and integration. These properties are summarised below.
Linearity of definite integrals
Just as for differentiation, when a function is multiplied by a constant $k$k and integrated, it is is the same as multiplying the integral of the function by the constant. This can simplify integral calculations.
$\int_a^b\ kf(x)\ dx=k\int_a^b\ f(x)\ dx$∫ba kf(x) dx=k∫ba f(x) dx
Practice question
Question 3
Given that $\int_{-2}^7f\left(x\right)dx=2$∫7−2f(x)dx=2, determine $\int_{-2}^75f\left(x\right)dx$∫7−25f(x)dx.
Additivity of definite integrals
If a definite integral is the sum or difference of two functions, then this is the same as adding or subtracting the individual integral of each function over the interval.
$\int_a^b\ f(x)+g(x)\ dx=\int_a^b\ f(x)+\int_a^b\ g(x)\ dx$∫ba f(x)+g(x) dx=∫ba f(x)+∫ba g(x) dx
Practice question
Question 4
Consider the function $f\left(x\right)$f(x) where $\int_{-1}^6f\left(x\right)dx=3$∫6−1f(x)dx=3.
Determine $\int_{-1}^6\left(9f\left(x\right)-2\right)dx$∫6−1(9f(x)−2)dx.
Reversal of an interval
If we reverse the interval of a definite integral, that is instead of integrating from $a$a to $b$b we integrate from $b$b to $a$a we will get the negative of the original integral.
$\int_a^b\ f(x)\ dx=-\int_b^a\ f(x)\ dx$∫ba f(x) dx=−∫ab f(x) dx
Practice questions
Question 5
Suppose $\int_4^6f\left(x\right)dx=3$∫64f(x)dx=3.
Find the value of $\int_6^4f\left(x\right)dx$∫46f(x)dx.
Find the value of $\int_4^63f\left(x\right)dx$∫643f(x)dx.
Find the value of $\int_4^6\left(f\left(x\right)+x\right)dx$∫64(f(x)+x)dx.
Odd/even functions
We can make use of the symmetrical properties of functions, including odd/even functions to simplify and more easily solve integrals.
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For an odd function $f(x)$f(x):
$\int_{-a}^a\ f(x)\ dx=0$∫a−a f(x) dx=0
For an even function $f(x)$f(x):
$\int_{-a}^a\ f(x)\ dx=2\ \int_0^a\ f(x)\ dx$∫a−a f(x) dx=2 ∫a0 f(x) dx
Practice questions
Question 6
Consider the function $y=x^3$y=x3.
Evaluate the integral $\int_0^2x^3dx$∫20x3dx.
Recognising that this is an odd function, write down the answer to $\int_{-2}^2x^3dx$∫2−2x3dx.
Using part (a), write down the area bounded by the function $y=x^3$y=x3, the $x$x-axis and the lines $x=-2$x=−2 and $x=2$x=2.
Splitting an interval
If we split an integral over a closed interval $[a,b]$[a,b] at the point $c$c where $c$c is within the closed interval, the sum of the integral of the parts equals the original integral.
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$\int_a^b\ f(x)\ dx=\int_a^c\ f(x)\ dx+\int_c^b\ f(x)\ dx$∫ba f(x) dx=∫ca f(x) dx+∫bc f(x) dx for $a
Practice questions
Question 7
Suppose $\int_{-1}^2f\left(x\right)dx=4$∫2−1f(x)dx=4 and $\int_2^8f\left(x\right)dx=8$∫82f(x)dx=8.
Find the value of $\int_{-1}^8f\left(x\right)dx$∫8−1f(x)dx.
Find the value of $\int_8^{-1}f\left(x\right)dx$∫−18f(x)dx.
Find the value of $\int_{-1}^22f\left(x\right)dx+\int_2^83f\left(x\right)dx$∫2−12f(x)dx+∫823f(x)dx.