Question

Can someone please help me integrate these problems attached?

Answer 1

Cinnnnn! Where's the attachment lady? c:

Answer 2

sorry lol

Answer 3

can you see it

Answer 4

there is an s \[\int\limits_{-14}^{1}s \left| 4-s^2 \right|\]

Answer 5

Oh oh oh, didn't see that >.< ty

Answer 6

sorry 49 still typo

Answer 7

\[\Large\rm \int\limits\limits_{-14}^1s\left|49-s^2\right|ds\]

Answer 8

correct

Answer 9

Lemme see if this makes sense....
\[\Large\rm \left|49-s^2\right|=\left|(7-s)(7+s)\right|\]
\[\Large\rm =\cases{-(7-s)(7+s), &s < -7\\ \rm \quad(7-s)(7+s), &-7 < s < 7}\]Do you understand what I'm doing with the pieces here?
Ahhh I can't.... words.... blah

Answer 10

yes you broke it down because it is an absolute value

Answer 11

So we have to rewrite our integral as the sum of two integrals.
We need to break up the boundary at -7

Answer 12

ok

Answer 13

would the bounds be -14 to -7 and -7 to 1

Answer 14

Yes those bounds sound correct c:

Answer 15

ok and then we integrate

Answer 16

\[\Large\rm \int\limits\limits_{-14}^{-7}-s(49-s^2)ds+\int\limits_{-7}^1 s(49-s^2)ds\]So I wrote these back as squares because we'll want to apply a substitution.
Easier to do so without factoring it.

The factoring was just to see where the bad spots were in our absolute.

Answer 17

ok I see question how did you know to split it at -7?

Answer 18

Oh boy.. how to put this into words.. ummm

Answer 19

If you look at the function y=x,

Answer 20

uhhhm

Answer 21

And take the absolute value of x,
what you're really doing is,

applying a negative to the x when x is negative.
That's what flips the branch up to being positive on the left side.
That's what the absolute function does, it applies a negative if we would be negative, changing it to positive in the end.

Answer 22

So when we have a factor of |7+s|,
when we're below -7, the inside is negative, so the absolute function tells us to apply a negative

Answer 23

Bahh this is the worst explanation >.<

Answer 24

lol

Answer 25

Yes I understand that, but in this equation would you just look for where the function became negative and that would be your bound

Answer 26

no it was good