Question

Find the area of the region between the graphs of f and g if x is restricted to the given interval

f(x) = 6 - 3x^2; g(x) = 3x; [0,2] Find the area of the region between the graphs of f and g if x is restricted to the given interval

f(x) = 6 - 3x^2; g(x) = 3x; [0,2] @Mathematics

Answer 1

find the intersection first

Answer 2

The intersections are -2 and 1.

Answer 3

\[\int\limits_{a}^{b}(\text{top function}-\text{bottom function})dx\]

Answer 4

Yes...I know that. :P I'm getting the wrong answer every time... :(

Answer 5

\[\int\limits_{-2}^{1} (6 - 3x^2 - x)dx\]

Answer 6

9?

Answer 7

Yea, that's the right answer, but I can't seem to get it?

Answer 8

note the set you are integrating over... [0,2]

Answer 9

you will have two integrals

Answer 10

That's what I've tried. However, I'd continually get something along the lines of 6-(8/3) and that kind of stuff.

Answer 11

integrate from 0 to 1 then from 1 to 2

Answer 12

Oh! Did NOT know that! Thanks. I should have, considering the graph makes an 'X' between 0 and 2 (with one graph increasing and another decreasing). I've sat here stumped for half an hour...jesus...thanks mate!

Answer 13

no problem

Answer 14

With the intersection occurring at 1 of course.

Answer 15

yes

Answer 16

that is why we split the integrals there

Answer 17

Thanks. I'll try it out and will give a report back in about 5 minutes. You've been given your well-earned "Good Answer" :P

Answer 18

Sorry, but I'm lost about one thing. How do I "split" it when I've only got two functions (thus only one upper and one lower limit?)

Answer 19

\[\int\limits_{0}^{1} (6 - 3x^2 - 3x)dx+\int\limits_{1}^{2} (3x-(6 - 3x^2) )dx\]

Answer 20

Wait, so you flip the "upper" limit even though it's the same function as the one that was classified as "lower" limit before reaching 1?

Answer 21

on [0,1] 6-3x^2 is the upper function and 3x is the lower
on [1,2] 3x is the upper function and 6-3x^2 is the lower