Question

a)
f(x) = 4x - 9. Find K so that (integral of (f(t))dt + k) [-1, x] = the integral of f(t) dt from [3,x]

b) find d/dx the integral f(t) dt from [-1,x]

Answer 1

looks pretty straightforward... have u tried anything yet ? :)

Answer 2

i have, but i dont know exactly what to do

i tried plugging in x/-x and than taking the derivative, but i didnt know how that would help me find k

Answer 3

you have this for part

\[\int\limits_{-1}^x f(t) dt + k = \int \limits_3^xf(t)dt\]

?

Answer 4

ya

Answer 5

familiar with below property ?
\[\int\limits_a^c f(t) dt = \int\limits_a^b f(t) dt + \int\limits_b^cf(t) dt\]

Answer 6

ya

Answer 7

Using that property you can split the given equation

\[\int\limits_{-1}^x f(t) dt + k = \int \limits_3^xf(t)dt \]

like below

\[\int\limits_{-1}^3 f(t) dt + \int\limits_{3}^x f(t) dt + k = \int \limits_3^xf(t)dt\]

yes ?

Answer 8

wait, why isnt the bottom of the right side of the equation -1?

Answer 9

why do u think it has to be -1 ?

Answer 10

we are splitting ` -1 to x` into `-1 to 3` and `3 to x`

Answer 11

would the 3 on top and bottom cancel out?

Answer 12

nope, but u can cancel something really big.. look closer

Answer 13

\[\int\limits_{-1}^3 f(t) dt + \color{red}{\int\limits_{3}^x f(t) dt} + k = \color{red}{\int \limits_3^xf(t)dt}\]

Answer 14

im confused

Answer 15

shouldnt the intervals be [-1,x] and what do you mean by something big is canceled out

Answer 16

[-1, x] changed to [-1, 3] + [3, x] using the property i gave you earlier

Answer 17

ya i understand that, but what you wrote with the intergral signs were different

Answer 18

im not sure what difference are u seeing :/

Answer 19

u put the intervals of [-1,3] + [3,x] = [3,x]

Answer 20

nope

Answer 21

you're given the equation
\[\int\limits_{-1}^x f(t) dt + k = \int \limits_3^xf(t)dt\]

Answer 22

splitting the integral on left side as [-1, 3] + [3, x] you get :

\[\int\limits_{-1}^3 f(t) dt + \int\limits_{3}^x f(t) dt + k = \int \limits_3^xf(t)dt\]

Answer 23

oooo i get that now

Answer 24

okay good :)

Answer 25

it actually becomes [-1,x] but that = [3,x] from the original equation

is that right?

Answer 26

yes

Answer 27

how do i use this to find k?

Answer 28

\[\int\limits_{-1}^3 f(t) dt + \color{Red}{\int\limits_{3}^x f(t) dt} + k = \color{Red}{\int \limits_3^xf(t)dt}\]
cancel that

Answer 29

\[\int\limits_{-1}^3 f(t) dt + k = 0\]

plugin the value of f(t) and evaluat ethe definite integral

Answer 30

\[\int\limits_{-1}^3 (4t-9)~ dt + k = 0\]

Answer 31

\[k = -\int\limits_{-1}^3 (4t-9)~ dt \]

Answer 32

evaluate

Answer 33

aite, now how do i do b?

Answer 34

what do u get for k ?

Answer 35

-16

Answer 36

my calc says 20, did i mess up?

Answer 37

nvm, forgot to take the anti-derivative

Answer 38

20 is right

Answer 39

http://www.wolframalpha.com/input/?i=+-%5Cint%5Climits_%7B-1%7D%5E3++%284t-9%29+dt

Answer 40

forgot to take anti-derivative , so i just plugged into 4x-9

Answer 41

yep i now got 20

Answer 42

sounds great!

Answer 43

for part b you want to compute \[\large \dfrac{d}{dx}\int\limits_{-1}^x f(t) dt \]

Answer 44

Say \(\int f(x) dx = F(x) + C \implies f(x) = F'(x)\) by fundamental thm of calculus

\[\large \begin{align} \dfrac{d}{dx}\int\limits_{-1}^x f(t) dt &= \dfrac{d}{dx}\left[F(t) \right]_{-1}^x\\~\\
&=\dfrac{d}{dx}\left[F(x) - F(-1)\right]\\~\\
&=\dfrac{d}{dx}F(x) - \dfrac{d}{dx} F(-1)\\~\\
&= f(x) - 0\\~\\
&=4x-9
\end{align}\]

Answer 45

Notice that `F(-1)` is the value of function evaluated at x=-1 which will be a `constant`. so its derivative will be 0.

Answer 46

thank you, you helped me clear the only problem on the test review that i was stuck on