Question

Please Help! Will FAN AND MEDAL!

Use the graph of f(t) = 2t + 3 on the interval [−3, 6] to write the function F(x), where f of x equals the integral from 3 to x of f of t dt.

a. F(x) = 2x2 + 6x
b. F(x) = 2x + 3
c. F(x) = x2 + 3x + 54
d. F(x) = x2 + 3x − 18

Answer 1

Can you please @jim_thompson5910 @agent0smith @ganeshie8

Answer 2

`Use the graph of f(t) = 2t + 3 on the interval [−3, 6] to write the function F(x), where f of x equals the integral from 3 to x of f of t dt.`

so

\[\Large f(x) = \int_{3}^{x}f(t)dt\]

\[\Large f(x) = \int_{3}^{x}(2t+3)dt\]

Answer 3

You'll use the fundamental theorem of calculus

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

where

g(x) = antiderivative of f(x)

Answer 4

Okay so I have to plug in the second problem and find the anti-derivative

Answer 5

what is the antiderivative of `2t+3` ?

Answer 6

t^2+3t+C

Answer 7

yep so let's make

g(t) = t^2+3t+C

Answer 8

now you need to compute g(x) and g(3) then subtract

g(x) - g(3) = ??

Answer 9

-3g

Answer 10

g(t) = t^2+3t+C

g(x) = x^2+3x+C ... replace every t with x

Answer 11

g(t) = t^2+3t+C

g(3) = (3)^2+3(3)+C ... replace every t with 3

g(3) = 9 + 9 + C

g(3) = 18 + C

Answer 12

so...

g(x) - g(3) = [x^2+3x+C] - [18 + C] = ???

Answer 13

tell me what you get or if you get stuck

Answer 14

\[x^2+3x-18\] @jim_thompson5910

Answer 15

correct. Nice work

Answer 16

Can you help me with one more? @jim_thompson5910