# Thread:

1. ## Area between to functions

Ok, so I was sick one lecture and couldn't make it but now I'm stuck on my homework assignment and cannot reach anybody to fill me in on how to complete certain problems (and our book is just awful in giving examples.)

I've figured out how to do problems with just the two functions themselves [ie: f(x) = 5-x^2 ; g(x) = x^2-3)] but I don't understand what to do when two x values are already given.

Problem: f(x) = 5-x^2 ; g(x) = x^2-3 ; x = 0, x = 4

The answer is 32. I keep getting stuck when I reduce the equation to -2x^3/3 + 8x (4 to 0)

2. Originally Posted by some_nerdy_guy
Ok, so I was sick one lecture and couldn't make it but now I'm stuck on my homework assignment and cannot reach anybody to fill me in on how to complete certain problems (and our book is just awful in giving examples.)

I've figured out how to do problems with just the two functions themselves [ie: f(x) = 5-x^2 ; g(x) = x^2-3)] but I don't understand what to do when two x values are already given.

Problem: f(x) = 5-x^2 ; g(x) = x^2-3 ; x = 0, x = 4

The answer is 32. I keep getting stuck when I reduce the equation to -2x^3/3 + 8x (4 to 0)
did you sketch the graphs of f and g ?

$A = \int_0^2 f(x) - g(x) \, dx + \int_2^4 g(x) - f(x) \, dx$

3. ## Interesting interpretation of bounded area

The 2 parabolas intersect at 2. From 0 to 2, the 5-x^2 parabola is on top of the other one. From 2 to 4 it's the other parabola that's on top. The question is asking for the area bounded by the 2 curves, so you need to split the problem into 2 parts, one integration from 0 to 2 and the other from 2 to 4. That gives you 32. Make sure you flip the sign of the integrand, otherwise the signed area will come out wrong.

4. Thanks a bunch...the sketching the graph part went over my head.
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