Question

Question about FCT?

Answer 1

I so far found the antiderivatives to get g(x).

Answer 2

I think you meant "FTC" for "fundamental theorem of calculus"

Answer 3

you're basically using FTC part 1, which says

If \[\Large g(x) = \int_{0}^{x}f(t)dt\]
then

\[\Large g'(x) = f(x)\]

Answer 4

it's possible that the endpoints of the intervals change from closed to open but I'm not 100% sure on that

Answer 5

oh nvm, it says nothing about finding the derivative of g

Answer 6

Okay

Answer 7

but yes you'll find the antiderivatives and use the second part of the FTC, which is

\[\Large \int_{a}^{b}f(x)dx = F(b) - F(a)\]

where F ' = f

Answer 8

okay, so far I've got g(x)= 0 if x<0, 1/2 * x^2 if 0<x<1, 2x-1.2*x^2 if 1<x<2, 0 if x>2

Answer 9

However, I'm just confused on if I should add a c two the second and third part of the piece wise function...

Answer 10

Thank you for helping me so far :)

Answer 11

you should have
\[\LARGE g(x) = \begin{cases}
C \ \text{ if } x < 0 \\
\frac{x^2}{2} \ \text{ if } 0 \le x \le 1 \\
2x-\frac{x^2}{2} \ \text{ if } 1 < x \le 2 \\
D \ \text{ if } x > 2 \\
\end{cases}
\]

C and D are constants

Answer 12

would it be x^2/2 + C?

Answer 13

oh right, forgot about the middle constants, let me redo

Answer 14

\[
\LARGE g(x) = \begin{cases}
C_{1} \ \text{ if } x < 0 \\
\frac{x^2}{2}+C_{2} \ \text{ if } 0 \le x \le 1 \\
2x-\frac{x^2}{2}+C_{3} \ \text{ if } 1 < x \le 2 \\
C_{4} \ \text{ if } x > 2 \\
\end{cases}
\]

C1 through C4 are constants

Answer 15

Alrighty, now how would we graph this? Can I state that C can equal zero and this will not affect the shape of the graph but only the vertical location of the graph?

Answer 16

although, when you use FTC part 2, I think the middle constants C2 and C3 will subtract and cancel out

the same may happen with C1 and C4. I'm not sure about that though

Answer 17

How does it cancel out?

Answer 18

if so, then you'd be left with

\[
\LARGE g(x) = \begin{cases}
0 \ \text{ if } x < 0 \\
\frac{x^2}{2} \ \text{ if } 0 \le x \le 1 \\
2x-\frac{x^2}{2} \ \text{ if } 1 < x \le 2 \\
0 \ \text{ if } x > 2 \\
\end{cases}
\]

Answer 19

well let's say that f(x) = x

the antiderivative is F(x) = (x^2)/2 + C

then use FTC part 2 to evaluate. Using FTC 2 means we use this property

\[\Large \int_{a}^{b}f(x)dx = F(b) - F(a)\]

Answer 20

when we get to the F(b) - F(a) part, the constants cancel (since you're subtracting C1 from itself, for instance)

Answer 21

I'm still not understanding... So we subtract the different parts of the piece wise function? Could you maybe give me an example please ? :)

Answer 22

let's use f(x) = x as an example

Answer 23

Jim, would the equations in g(x) be considered integrals? The problem states to not include them in g(x) *worry face*

Answer 24

actually let f(t) = t

\[\Large \int_{0}^{x}f(t)dt\]

\[\Large \int_{0}^{x}tdt\]

\[\Large \left. \frac{t^2}{2}+C\right]_{0}^{x}\]

\[\Large \left[\frac{x^2}{2}+C\right]-\left[\frac{0^2}{2}+C\right]\]

\[\Large \frac{x^2}{2}+C-\frac{0^2}{2}-C\]

\[\Large \frac{x^2}{2}\]

Answer 25

so hopefully you can see how

\[\Large \int_{0}^{x}f(t)dt = \Large \frac{x^2}{2}\]

and how the constant canceled

Answer 26

Yes, i see now!

Answer 27

so that's what I think is happening with each sub-interval (and each antiderivative)

Answer 28

Then you simply just graph f(x) and g(x) on those intervals?

Answer 29

yes correct

Answer 30

if you want, you can plot points for g(x)

to generate points for g(x), you simply record the area under the curve from 0 to x (under f(x))

Answer 31

Wait so we would be follow the equation we made for g(x) to make the graph of it?

Answer 32

that is another alternative, yes

Answer 33

Alright, thank you Jim.

Answer 34

you're welcome