If you know that the area of the shaded region is 40 square units, which of the following can you evaluate?
A summary of the integrals and their values is provided below:
|\(\int_2^6 (f(x) + 5) \ dx\)||\(60\)|
|\(\int_2^6 (f(x) - 3) \ dx\)||\(28\)|
|\(\int_0^4 f(x + 2) \ dx\)||\(40\)|
|\(\int_2^6 f(x + 2) \ dx\)||?|
|\(\int_4^8 f(x + 2) \ dx\)||?|
|\(\int_2^6 -f(x) \ dx\)||\(-40\)|
|\(\int_2^6 f(-x) \ dx\)||?|
We can consider each in turn to understand how these values can be calculated.
Can you find the value of the constant \(k\) for which
\[\int_2^6 (f(x) + k) \ dx = 0\]
For the integral to equal \(0\), we need to introduce a negative component. In order to do this we need to translate the curve ‘downwards’ such that we achieve a situation like this, in which the area under the curve located ‘above’ the \(x\)-axis is equal to that ‘below’ the \(x\)-axis. This in turn means that \(k\) will be negative.
We have a maximum of three portions to consider. In the diagram shown, the area of ‘portion’ \(B\) must be equal to the sum of the areas of ‘portions’ \(A\) and \(C\).
From now on we will use \(A\) to represent the area of ‘portion’ \(A\), \(B\) to represent the area of ‘portion’ \(B\), etc.)
The value of the integral \(\int_2^6 (f(x) + k) \ dx\) will be equal to \(B - (A + C)\).
The area of ‘portion’ \(D\) can be written in two ways: \(4 \vert k \vert - (A+C)\) or \(40 - B\).
In this approach we have chosen to think about areas of ‘portions’ of the diagram. This means that even though \(k\) must be negative (by our own initial reasoning), we will use the magnitude of \(k\) (its positive value) for area calculations.
\(\vert k \vert\) denotes the magnitude of \(k\).
Equating these two expressions we have \(D = 4 \vert k \vert - (A+C) = 40 - B\) and rearranging we find that \[B - (A+C) = 40 - 4 \vert k \vert.\]
In this instance we want the integral to equal \(0\) so we can write \[0 = 40 - 4 \vert k \vert\] and therefore \[\vert k \vert = 10.\]
We know that \(k\) is in fact negative given the set up of our diagram and therefore \(\int_2^6 f(x) - 10 \ dx = 0\).
As in our approach to some of the original integrals, it may be worth noting that an algebraic approach here will also work. Re-writing the integral as \[\int_2^6 (f(x) + k) \ dx = \int_2^6 f(x) \ dx + \int_2^6 k \ dx = 0\] enables us to see that \(\int_2^6 k \ dx = -40\).
Proceeding with the integration we then find that \(4k = -40\) and therefore \(k=-10\) as before.