Graph 7 is similar to graph 8, how are they different?

given graph of f(x),sketch a graph of A(x), the area under function f(x)

May 31, 2018

Graph 7: $0 \le x \le a$: $A \left(x\right) = \frac{1}{2} k \left({a}^{2} - {\left(x - a\right)}^{2}\right)$
$x > a$: $A \left(x\right) = \frac{1}{2} k \left({a}^{2} + {\left(x - a\right)}^{2}\right)$
Graph 8: $A \left(x\right) = \frac{1}{3} k \left({\left(x - a\right)}^{3} + {a}^{3}\right)$

Explanation:

I am in doubt if you are supposed to actually work out the function for A(x) in both instances, but having the actual function makes it easier to discuss the differences.

The main difference with the two graphs is that A(x) for graph 8 is rising sharper than what graph 7 does. This is due to the fact that the area for graph 7 is a second degree function (a combination of two half parabolas, one with the top when x=a, the other with bottom when x=a), while graph 8 is a third degree function, which, therefore, changes the slope much faster.

First graph 7: $f \left(x\right) = k | x - a |$, which has its bottom at
$f \left(0\right) = a$. ($x \ge 0$)

Using parallellograms we can show that the area function is on the form
$0 \le x \le a$: $A \left(x\right) = \frac{1}{2} k \left({a}^{2} - {\left(x - a\right)}^{2}\right)$
$x > a$: $A \left(x\right) = \frac{1}{2} k \left({a}^{2} + {\left(x - a\right)}^{2}\right)$

We must consider $0 \le x \le a$ and $x > a$ separately since the graph switches from a downward slope to an upward slope when x=a.

For $0 \le x \le a$ we have:

It should be apparent that the figure ABCE is a parallellogram, which for the value x has the height $x \left(= A D\right)$ and bottom and top width respectively $A B = f \left(0\right) = k a$, and $C D = k \left(a - x\right)$ (since $a > x$).
The area, therefore, is
$A \left(x\right) = \frac{1}{2} \left(A B + C D\right) x = \frac{1}{2} \left(f \left(0\right) + f \left(x\right)\right) x = \frac{1}{2} k x \left(2 a - x\right)$

Similarly we will find
$A \left(x\right) = \frac{1}{2} k \left({a}^{2} + {\left(x - a\right)}^{2}\right)$ when $x > a$

Graph 8:
This is a parabola on the form of $f \left(x\right) = k {\left(x - a\right)}^{2}$

If we slize the area underneath the graph in n pieces, we get

Using the sum of the parallellograms as an approximation of the area we can show that as $n \to \infty$,
$A \left(x\right) = \frac{1}{3} k \left({\left(x - a\right)}^{3} + {a}^{3}\right)$

As this is getting overly long, I will stop there and leave the rest to you.