Question

\[\int\limits_0^\infty f(x)e^{-sx}\text dx\]
\[f(x) =\begin{cases}x^2 & 0\leq x \leq2\\4 & 2 13 years ago

Answer 1

\[\int\limits_0^\infty f(x)e^{-sx}\text dx\]\[=\int\limits_0^2 x^2e^{-sx}\text dx+\int\limits_2^\infty 4e^{-sx}\text dx\]\[=\left.\frac{x^2e^{-sx}}{-s}\right|_0^2+\frac 2s\int_0^2 xe^{-sx}\text dx+\left.\frac{4e^{-sx}}{-s}\right|_2^\infty\]\[=\frac{4e^{-2s}}{-s}+\frac 2s\left[\left.\frac{xe^{-sx}}{-s}\right|_0^2+\frac1s\int_0^2e^{-sx}\text dx\right]+\frac{4e^{-2s}}{s}\]\[=\frac{4e^{-2s}}{-s}+\frac 2s\left[\frac{2e^{-2s}}{-s}+\left.\frac{e^{-sx }}{-s^2} \right|_0^2\right]+\frac{4e^{-2s}}{s}\]\[=\frac{4e^{-2s}}{-s}+\frac 2s\left[\frac{2e^{-2s}}{-s}+\frac{e^{-2s}}{-s^2}+\frac{1}{s^2}\right]+\frac{4e^{-2s}}{s}\]\[=-\frac{4e^{-2s}}{s}-\frac{4e^{-2s}}{s^2}-\frac{2e^{-2s}}{s^3}+\frac{2}{s^3}+\frac{4e^{-2s}}{s}\]\[=\frac{2-2e^{-2s}(1+2s+2s^2)}{s^3}+\frac{4e^{-2s}}{s}\]\[=\frac{2-2e^{-2s}(1+2s)}{s^3}\]

Answer 2

have i made any mistakes?

Answer 3

laplace transform?

Answer 4

is transformation necessary?

Answer 5

I thought that was what you were doing?

Answer 6

laplace transformation might be an easier way ,

Answer 7

yep, with step function

Answer 8

x^2 u(t)+ (-x^2+4)u(t-2)

Answer 9

are the solutions consistant?

Answer 10

I think so
http://www.wolframalpha.com/input/?i=laplace+transform+of+t%5E2+heaviside%28t%29%2B+%28-t%5E2%2B4%29heaviside%28t-2%29

Answer 11

thanks