The fundamental theorem of Calculus. Help?
@zepdrix @phi @dan815
Ooo the 4th one is pretty fun :)
ok, so you want to integrate y over the interval, can you write the experssion
Is my 1st one right?
yes
For number 2 I got 1/3 is that right too?
yes c:
ok so number 3 and 4 I am confused
I know for 4 they are looking for the inverse but how do i get it?
Inverse? :o Derivative silly! :)
ohhhhhhh my bad nvm give me a second
You have to apply the Fundamental Theorem of Calculus, Part 1:\[\large\rm \frac{d}{dx}\int\limits_c^x g(t)dt=g(x)\] Actually it might be useful to use BOTH parts of the theorem for this problem.
I got 18x^3 for the derivative of it.
but that isnt a choice =C
For nuber 3 I got B. and number 4 I don't know.
I got 18x^3 when I integrated but 0 for the derivative...**
for 4 you have to experss F as a function of x, not as intergration
I got 18x^3 when i integrated bu 0 as the derivative**
This is what we're trying to figure out problem 4,\[\large\rm \frac{d}{dx}\int\limits_{-3x}^{3x}s^2ds\]Try to think of it more generally like this,\[\large\rm \frac{d}{dx}\int\limits_{-3x}^{3x}f(s)ds\]You anti-differentiate to get some function, let's call it F(s)
\[\large\rm \frac{d}{dx}\int\limits\limits_{-3x}^{3x}f(s)ds\quad=\frac{d}{dx}F(s)|_{-3x}^{3x}\]
Plug in your limits,\[\large\rm \frac{d}{dx}\left[F(3x)-F(-3x)\right]\]and now take a derivative
Big F turns back into the original function, little f, but we have some chain rule going on, right?\[\large\rm =3f(3x)--3f(-3x)\]
yeah
little f is the thing we started with, s^2 in this case, So now we have the 3x and -3x being squared, replacing the s value,\[\large\rm =3(3x)^2+3(-3x)^2\]And simplify! :)
It's a weird little trick. If you got confused somewhere in that process, I wouldn't be surprised.
54x^2 so c. Now number 3.
so b*
I'm going to close this one and open a new one
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