Question

FTC part 2 given a graph

Answer 1

I don't know what to do with this one. Should I just do what you taught me earlier?

Answer 2

#7 and #8 are area problems

Answer 3

#9 and #10 use FTC2

Answer 4

for #7 :

\[\large F(0) = \int \limits_{-6}^0 f(t) dt\]

Answer 5

find the area under curve f(t)
between x = -6 and 0

Answer 6

ok I can do 7 and 8

Answer 7

for 9 and 10 do I set it up like you showed me before

Answer 8

for 7 and 8 careful about the negative ares *

Answer 9

when the area is below x axis, integral is negative of Area

Answer 10

Thanks just notice that

Answer 11

7 is -2pi +2
8 is -2pi +(1/2)

Answer 12

not sure if I am doing 9 correct

Answer 13

I got -8 for a solution but I did not use the graph at all. So I know I am incorrect

Answer 14

for #9 and #10 :
\[\large F(x)= \int \limits_{-6}^{2x}f(t) dt\]
\[\large F'(x) = \dfrac{d}{dx} \int \limits_{-6}^{2x}f(t) dt\]
\[\large = f(2x) (2x)'\]
\[\large = 2f(2x) \]

Answer 15

So,

\[\large F'(x) = 2f(2x) \]

Answer 16

oh, I forgot to write f(2x)

Answer 17

just plugin x = -2, for #9

Answer 18

is it -4?

Answer 19

Yes !

Answer 20

what about 11? 2nd derivative
do I just take the derivative of the above

Answer 21

we have : \[\large F'(x) = 2f(2x) \]

Answer 22

differentiate both sides with respect to x

Answer 23

\[\large F''(x) = 2f'(2x) \]

Answer 24

plugin x = 0

Answer 25

shouldn't that be a 4 in front

Answer 26

\[F"(x)=4f ' (2x)\]

Answer 27

thats a very good question, let me think a bit...

Answer 28

You're right ! we need to use chain rule and ithas to be 4f'(2x)

Answer 29

cool, and my solution to 11 is 4

Answer 30

looks good :)

Answer 31

thanks you make it look so easy and give me hope

Answer 32

np :)