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:
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\(\displaystyle g(x)\) 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: \(\displaystyle \:g(x)\:=\:\begin{Bmatrix}0\;\;\;\;\;\;\;\;x\,<\,0 \\ \frac{5}{2}x^2\;\;\;\;\;0\,\leq\,x\,\leq\,6 \\ 90\,+\,60x-\frac{5}{2}x^2\;\;\;6\,<\,x\,\leq\,12 \\ 180 \;\;\;\;\;\;\;x\,>\,12\end{Bmatrix}\)