Question

Let
g(x) = integral 0-x of f(t) dt, where
f(x) is the piecewise function below.

http://i45.tinypic.com/2crplw5.jpg

Answer 1

Okay.... \[12x-\frac{ x ^{2} }{ 2 }\] for 3 < x < 8 interval

Answer 2

I tried that...no bueno

Answer 3

that is because you ignored the constant

Answer 4

don't forget that at \(x=3\) you should have \(F(3)=0\)

Answer 5

clear or no?

Answer 6

i'm checkin now

Answer 7

no good

Answer 8

it is \[12x-\frac{ x ^{2} }{ 2 }-(12\times 3-\frac{ 3 ^{2} }{ 2 })\]

Answer 9

\[12x-\frac{x^2}{2}-\frac{63}{2}\]

Answer 10

or maybe write in standard form

Answer 11

\[-\frac{x^2}{2}+12x-\frac{63}{2}\]

Answer 12

oh wait, that is totally wrong, sorry
you want to add the accumulated value up until \(x=3\)

Answer 13

up until \(x=3\) you get \(\frac{3^3}{3}=9\) so your function should be \(9\) at \(x=3\) not 0 as i said above

Answer 14

\[-\frac{x^2}{2}+12x+c\] replace \(x\) by \(3\) set the result equal 9 to get \(c\)

Answer 15

63/2 for c?

Answer 16

\[-\frac{3^2}{2}+36+c=9\] \[c=-\frac{45}{2}\] i think should work
try \[-\frac{x^2}{2}+12x-\frac{45}{2}\]

Answer 17

no that was my mistake
it should be equal to 9 and \(x=3\) not 0

Answer 18

you are a genius!

Answer 19

well thanks, but not hardly
messed up mightily the first time