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

## Homework Statement

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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.

## Homework 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
$Area = -\int_a^b f(x) dx$

## The Attempt at a Solution

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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:

RUber
Homework Helper
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.

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.