Question

Let g(x) = integration from 0 to x of f(t) dt, where
f(x) is the piecewise function below.

f(x) = 0 x<0
x^2 0<(or equal to) x < 3
12-x 3<(or equal to) x < 10
2 x >(or equal to) 10

Given that f(x) is continuous, write g(x) as a piecewise function, in similar fashion to f(x).

Answer 1

One piece at a time:

\(\int\limits_{-\infty}^0 0\; dt = 0\)

\(\int\limits_0^3 t^{2}\;dt = What?\)

Answer 2

they set it up with the same intervals like fromx<0 then 0<x<3 etc.

Answer 3

so g(x)= ? for x<0 ....

Answer 4

g(x) = 0 for x < 0

Answer 5

Sorry, it should have been, for x < 0, \(g(x) = \int\limits_{-\infty}^{x}0\;dt\)

Likewise, for 0 < x < 3, we have \(g(x) = \int\limits_{0}^{x}t^{2}\;dt\)

With a little luck, it will be continuous at x = 0

Answer 6

what about for intervals 3<(or equal to) x < 10 and x >(or equal to) 10 ? would it be more than integration?

Answer 7

you have to add what you have accumulated so far

Answer 8

\(g(x)=0\) if \(x<0\)
\(g(x)=\frac{x^3}{3}\) if \(0\leq x<3\)

Answer 9

since \(g(3)=\frac{3^3}{3}=9\) we have on \(3<x<10\)
\[g(x)=12x-\frac{x^2}{2}+c\] and we find \(c\) by noting that \[g(3)=12\times 3-\frac{3^2}{2}+c=9\]

Answer 10

thank you :)