Area in between graphs, one graph partially below y=0

1. Dec 17, 2014

Keipi

1. The problem statement, all variables and given/known data

The problem is stated in dutch and dutch is my first language. I will try to translate it all as accurately as possible.

Imagine the following two functions: $f(x)=x^3-4x^2$ and $g(x)=2x^2$.
Algebraically calculate the area in between the two graphs.

2. Relevant equations

Integration:
$f(x)=x^a$ then $\int f(x)dx=\frac{1}{a+1} x^{a+1}+c$

Area underneath a graph where it's under y=0 in between x=a and x=b
$Area = -\int_a^b f(x) dx$

3. The attempt at a solution

I began by finding the two intersecting points of the graphs to determine for which x-values I needed to integrate.

$f(x)=g(x)$ so $x^3-4x^2=2x^2 \rightarrow x^3-6x^2=0 \rightarrow x^2(x-6)=0 \rightarrow x^2 = 0 or x-6 = 0$ thus x=0 and x=6 are the two points of intersection.

Afterwards I determined where $f(x)$ intersects with $y=0$, to see which part I needed to calculate separately.

$x^3-4x^2=0 \rightarrow x^3=4x^2 \rightarrow x=4$

The part of f(x) that is underneath the x-axis needs to be added to the area below g(x) since it's part of the enclosed area. The calculation for the area will then be:

$\int_0^6 g(x) dx +(-\int_0^4 f(x)dx) - \int_4^6 f(x) dx$
$[\frac{2}{3}x^3]_0^6 +(-[\frac{1}{4}x^4-\frac{4}{3}x^3]_0^4) - [\frac{1}{4}x^4-\frac{4}{3}x^3]_0^4$
$144+21\frac{1}{3}-58\frac{1}{3}=108$

The answer 108 corresponds with the book, so it either is a coincidence or I used a method that was right but different from the book. The book calculates it like this:

$\int_0^6 g(x)-f(x) dx \rightarrow \int_0^6 2x^2 - (x^3 - 4x^2) dx = \int_0^6 6x^2-x^3$
Integrating the above also results in the answer of 108.

The part I don't understand is how their method is correct. f(x) in the domain of x=0 to x=4 is below the x-axis, as far as I know this would mean having to split it up and calculating it separately. The part between the x-axis and f(x) in the domain of x=0 to x=4 would be left out using their method. Or so I think.

The question then comes down to this: If the method used by the book is correct, why? Shouldn't the part below the x-axis be integrated seperately?

Not to be pessimistic or negative about the community on this forum (I don't know the community here, this is just a generalization of the average forum poster. Sorry in advance if I am horribly wrong.), I expect the answer to be 'ask your teacher for an explanation, that would be best.'. To avoid this, I felt I should explain the situation so I can get a proper answer quicker:

I do not have a teacher. I am doing this by myself, the only book I have is a summary book and the problem I listed above is from an exam given in previous years. The book I have, being a summary book, does not give any proper explanation for the statements it does so I mostly learn by trial and error and trying to understand what is going on. The reason I need the knowledge is because I have to take the exam myself so I can start my bachelor study in physics next college year.

I hope someone can explain this to me. If anything is unclear, please do ask. I have proof-read it but I can't guarantee that I got everything.

Last edited by a moderator: Jan 5, 2015
2. Dec 17, 2014

RUber

Both methods are good. You can just subtract the functions since as long as f is below g, the function h=g-f is positive. So, you do not need to split up h over the interval.

3. Dec 17, 2014

LCKurtz

To add to what RUber said, the formula for the area between two graphs between $a$ and $b$ is$$A = \int_a^b y_{upper}-y_{lower}~dx$$If the curves don't cross, there is no problem. Where the trouble usually comes for students is where they are calculating the area between a function $f(x)$ and the $x$ axis. If $f(x)$ goes negative on the interval the upper and lower curves ($f(x)$ and the $x$ axis) change. That's when you have to break the interval up.

4. Jan 5, 2015

Keipi

Sorry for the extremely late answer. Thanks for the help, after the answers I went to check again to see what I was missing, why I couldn't really see that it was both true.

After working through it again I noticed that taking the negative integral of the part underneath the x-axis will turn out positive, and then be added to the rest. Knowing that, it seemed logical to me that you could also just add it to the other integrals at the start to save time.

Thanks again. :)