Question

@ganeshie8 FTC

Answer 1

just when I thought I understood the concept, then I get these problems

Answer 2

38 is 4 using FTC

Answer 3

I have not clue how to do 39 or 40

Answer 4

what am I overlooking?

Answer 5

since the given graph itself is \(f'(x)\), integral gives area under that graph

Answer 6

Note that in the earlier question, we're given the graph of \(f(x)\)

Answer 7

yes you are correct

Answer 8

\(\int_0^7 f'(x) dx\) = area under the given graph

Answer 9

break it into triangles/rectangles and add up

Answer 10

which one you stuck at?

Answer 11

no I can redo 38
still don't know what to do with 39 and 40

Answer 12

we can use FTC for them

Answer 13

for 39, try below :

\(f(0) - f(-3) = \int_{-3}^0 f'(x) dx\)

Answer 14

ok got it, will be able to do 39 and 40 now.
38 is 9

Answer 15

\(f(3) - f(0) = \int_{0}^3 f'(x) dx\) *

Answer 16

f(3)=-6

Answer 17

f(7)=-7

Answer 18

thanks you are awesome :)

Answer 19

9 is correct for 38

Answer 20

how did u get f(3) = -6 ?

Answer 21

f(0)-f(3)= integral (sorry too lazy to type it) 0 to 3 f'(x)dx
-3-f(3)=3
-f(3)=6
f(3)=-6

Answer 22

did I do something incorrect?

Answer 23

\(f(3) - f(0) = \int_0^3 f'(x) dx\)

\(f(3) - f(0) = 3\)

\(f(3) -(-3) = 3\)

\(f(3) = 0\)

Answer 24

It is always : (end value) - (initial value)

Answer 25

omg maybe I have reached my limit of math today

Answer 26

ok I will fix 40

Answer 27

hahah same for #40 also :

\ok...

Answer 28

f(7)=5 I feel like the student who has been corrected by the teacher (once again)

Answer 29

All good teachers are good students too :)

Answer 30

lol

Answer 31

f(7) = 5 looks good !