# Thread: Area between the graphs

1. ## Area between the graphs

ok i got the parts where there is a number in the field but im way confused on what it wants for the other ones. Can you help?

Consider the area between the graphs and . This area can be computed in two different ways using integrals.

First of all it can be computed as a sum of two integrals
with the following values:

Alternatively this area can be computed as a single integral
with the following values:

Either way we find that the area is .

2. Originally Posted by amiv4
ok i got the parts where there is a number in the field but im way confused on what it wants for the other ones. Can you help?

Consider the area between the graphs and .
check this. did you really mean to type 1y like that?

3. its an online assignment so i just copied it straight from there and that is how they have it. I'm not sure why but they do

4. Originally Posted by amiv4
its an online assignment so i just copied it straight from there and that is how they have it. I'm not sure why but they do
i will say this. the area between two curves is given by $\displaystyle \int (\text{top function} - \text{bottom function})~dx$ or $\displaystyle \int (\text{right function} - \text{left function})~dy$

so draw the two curves and see what these functions are and the limits of integration they carry

5. k i got all the information on the first part, but then on the second part i only have the bottom limit of the integral which is -5 but idk wat the top limit is. Cuz i tried 11 and 20 and those dont work and idk wat else it could be

6. Originally Posted by amiv4
k i got all the information on the first part, but then on the second part i only have the bottom limit of the integral which is -5 but idk wat the top limit is. Cuz i tried 11 and 20 and those dont work and idk wat else it could be
you had to solve $\displaystyle 15 - y = y^2 - 5 \implies y^2 + y - 20 = 0 \implies (y + 5)(y - 4) = 0 \implies y = -5 \mbox{ or } y = 4$

7. oh ok i get it now. thanks