Food for thought

## Solution part 1

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:

Integral Value
$\int_2^6 (f(x) + 5) \ dx$ $60$
$\int_2^6 (f(x) - 3) \ dx$ $28$
$\int_0^4 f(x + 2) \ dx$ $40$
$\int_2^6 f(x + 2) \ dx$ ?
$\int_4^8 f(x + 2) \ dx$ ?
$\int_2^6 -f(x) \ dx$ $-40$
$\int_2^6 f(-x) \ dx$ ?

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.