Question

I get that its the second fundamental theorem... and I know how to do the problems just what the hell do they want me to answer for the part not done, i am on my last attempt.

Answer 1

how did you get F(3), F(5), F(7) correct if you didn't get F(x) correct? That's odd

Answer 2

i just dont get what they mean lol

Answer 3

what do you get for the antiderivative of -2/(t^3) ?

Answer 4

lagged excuse me:

\(\large\color{blue}{\displaystyle\int\limits_{a}^{b}f(x)~dx=~F(x)~{{\huge |}_{ a}^{b}}=F(b)-F(a)}\) where F'=f (we know that)


\(\large\color{blue}{\displaystyle\int\limits_{a}^{g(x)}s(t)~dt=~S(t)~{{\huge |}_{ a}^{g(x)}}=S(f(x))-S(a)}\)

Answer 5

your g(x) is x.
s(t) is -2/t^3
a is 3

Answer 6

would it be -2x/x^3+C i am not the best at this yet

Answer 7

\(\large\color{blue}{\displaystyle\int\limits_{}_{}-\frac{2}{t^3}~dt}\) find this

Answer 8

\(\large\color{blue}{\displaystyle\int\limits_{}_{}-2t^{-3}~dt~~~=~?}\)

Answer 9

(you can take the constant out and apply the power rule)

Answer 10

(2t^-2)/2

Answer 11

Correct, that simplifies to what ?

Answer 12

(cancel out the 2s)

Answer 13

t^-2

Answer 14

so i just plug in x and i got my answer alright

Answer 15

yes.
So we know that: \(\large\color{blue}{\displaystyle\int\limits_{}_{}-2t^{-3}~dt=t^{-2}}\)

almost, (to last reply)

Answer 16

disregard +C fopr now

Answer 17

\(\large\color{black}{\displaystyle\int\limits_{}_{}-2t^{-3}~dt=t^{-2}}\)

\(\large\color{black}{\displaystyle\int\limits_{3}^{x}-2t^{-3}~dt=t^{-2}{{\huge |}_{3}^{x}}}\)

Answer 18

making sense ?

Answer 19

you are plugging in x, correct, but you are also plugging in ?

Answer 20

3

Answer 21

yes.

Answer 22

so you are dong this:

\(\large\color{black}{\displaystyle\int\limits_{3}^{x}-2t^{-3}~dt=\frac{1}{t^2}~ {{\Huge |}_{3}^{x}}=\left[ \frac{1}{x^2} \right]-\left[ \frac{1}{3^2} \right]}\)

Answer 23

it wasn't right and i used all my submissions

Answer 24

what wasn't right?

Answer 25

this what I just said? if so you shall know that this is not the final answer.

Answer 26

x^-2

Answer 27

crap that is 1/x^2

Answer 28

-:(

Answer 29

oh well 3/4 points lol

Answer 30

when they are asking for F(x) do they want the evaluation of the integral, or just of the upper limit ?

Answer 31

idk the thing didn't give an answer afterwards

Answer 32

technically,

\(\large\color{black}{\displaystyle F(x)=\int\limits_{3}^{x}-2t^{-3}~dt=\frac{1}{t^2}~ {{\Huge |}_{3}^{x}}=\left[ \frac{1}{x^2} \right]-\left[ \frac{1}{3^2} \right]=\frac{1}{x^2}-\frac{1}{9}}\)

alternate forms,

\(\large\color{black}{\displaystyle\frac{9-x^2}{9x^2}}\)


\(\large\color{black}{\displaystyle\frac{(3-x)(3+x)}{9x^2}}\)

Answer 33

i think it was supposed to be (1/x^2)-(1/9)

Answer 34

the integral yes.
but the upper limit would just be 1/x^2,

Answer 35

yeah whatever hopefully it wont be just me who got confused and the teacher will give me a chance to redo the problem with different numbers

Answer 36

that is what I am getting confused about, do they want the upper limit or the integral (?)

Answer 37

anyway thank you solomon have a good night

Answer 38

yes, hopefully.
that is why I dislike multiple choice.

I just prove that I understand and teacher lets me pass by.

Answer 39

anyway bye