Find an expression...

nikchic5

Junior Member
Joined
Feb 16, 2006
Messages
106
Let
f(x)=
0 if x is less than 0
5x if 0 is less than or equal to x is less than or equal to 6
60-5x if 6 is less than x is less than or equal to 12
0 ix x is less than 12

and g(x)= integral from 0 to x f(t)dt. Find an expression for g(x) similar to the one for g(x)

Thanks so much for your help!
 
To be clear, is this the function:
\(\displaystyle \L
f(x) = \left\{ \begin{array}{c l}
0\quad &:&\quad x < 0 \\
5x\quad &:&\quad 0 \le x \le 6 \\
60 - 5x\quad &:&\quad 6 < x \le 12 \\
0\quad &:&\quad 12 < x \\
\end{array} \right.\)
 
Yes

Yes that is the correct function. Could you please help me??? Thanks so much!
 
Hello, nikchic5!

The last inequality must be: \(\displaystyle x\,>\,12\)


Let \(\displaystyle \,f(x)\:=\:\begin{Bmatrix}0\;\;\;\;\;\;x\,<\,0 \\ 5x\;\;0\,\leq \,x\,\leq\,6 \\ 60-5x\;\;6\,<\,x\,<\,\leq 12 \\ 0\;\;\;\;\;\;x\,>\,12\end{Bmatrix}\)

and \(\displaystyle \,g(x)\:=\L\int^{\;\;\;x}_0f(t)\,dt\)

Find an expression for \(\displaystyle g(x)\) similar to the one for \(\displaystyle f(x)\).
The graph of \(\displaystyle f(x)\) looks like this:
Code:
              |     (6,30)
              |       *
              |     *   *
              |   *       *
              | *           *
      - * * * * - - - + - - - * * * * -
              |       6      12
\(\displaystyle g(x)\) is the area under \(\displaystyle f(x).\)


For \(\displaystyle x\,<\,0\), the area is \(\displaystyle 0.\)
For \(\displaystyle 0\,\leq\,x\leq\,6\), the area is: \(\displaystyle \L\,\int_0^{\;\;\;x}\)\(\displaystyle 5t\,dt\;=\;\frac{5}{2}x^2\)

For \(\displaystyle 6\,<\,x\,\leq 12\), the area is: \(\displaystyle \L\,\int_0^{\;\;\;x}\)\(\displaystyle (60-5t)\,dt\:=\:60x\,-\,\frac{5}{2}x^2\)
Hence: \(\displaystyle \,g(x)\;=\;90\,+\,60x-\frac{5}{2}x^2\)


For \(\displaystyle x\,>\,12\), the area is \(\displaystyle 0.\)
Hence: \(\displaystyle \,g(x)\:=\:180\)

Therefore: \(\displaystyle \:g(x)\:=\:\begin{Bmatrix}0\;\;\;\;\;\;\;\;x\,<\,0 \\ \frac{5}{2}x^2\;\;\;\;\;0\,\leq\,x\,\leq\,6 \\ 90\,+\,60x-\frac{5}{2}x^2\;\;\;6\,<\,x\,\leq\,12 \\ 180 \;\;\;\;\;\;\;x\,>\,12\end{Bmatrix}\)
 
There is a small error in the g function above:
\(\displaystyle \L
6 < x \le 12\quad \Rightarrow \quad 90 + \int\limits_6^x {60 - 5tdt} = 60x - \frac{5}{2}x^2 - 180\)
 
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