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I understand that I need to convert to Heaviside functions and I think I'm doing that correctly, however my answer is not what it's supposed to be.

Below is one of the problems I am working on, with my workings and interpretation of what the graph should look like. Could someone please explain how I am meant to get my answer to look like that.

My work

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If $f(t)=g(t)u(t-t_0)$, then the Laplace Transform, $F(s)$, of $f(t)$ is

$$\begin{align} F(s)&=\int_0^\infty f(t)e^{-st}\,dt\\\\ &=\int_0^\infty g(t)u(t-t_0)e^{-st}\,dt \\\\ &\int_{t_0}^\infty g(t)e^{-st}\,dt \tag 1 \end{align}$$

Now, we enforce the substitution $t-t_0=t'$ into $(1)$ to obtain

$$F(s)=\int_0^\infty g(t'+t_0)e^{-s(t'+t_0)}\,dt' \tag 2$$

Using $(2)$ with $g(t)=t^2$ and $t_0=5$ yields

$$\begin{align} F(s)&=\int_0^\infty (t+5)^2e^{-s(t+5)}\,dt\\\\ &=e^{-5s}\int_0^\infty (t^2+10t+25) e^{-st}\,dt\\\\ &=e^{-5s}\left(\frac{2}{s^3}+\frac{10}{s^2}+\frac{25}{s}\right) \end{align}$$

Note the in general, if $F(s)$ is the Laplace transform of $f(t)$, then $e^{-st_0}F(s)$ is the Laplace Transform of $f(t-t_0)$. For the function $f(t)=t^2$, the Laplace transform is $\frac{2}{s^3}$.

Therefore, the Laplace transform of $(t-5)^2u(t-5)$ is $e^{-5s}\left(\frac{2}{s^3}\right)$. But then note that $(t-5)^2u(t-5)\ne t^2u(t-5)$. And this was, I believe, the issue with the OP.

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    I did it now using the laplace definition but I don't understand why method didn't work, could you please help me understand. – 2017-01-20
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    Sure. I've edited and provided a "walk-through." – 2017-01-20
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    I see where I went wrong, thank you so much for the assistance. – 2017-01-20
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    You're welcome! My pleasure. -Mark – 2017-01-20
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I think what you have learnt yet is $L[f(t-a).u(t-a)]=e^{-as}.L[f(t)]$ which is somewhat tricky to acheive in this problem and what you really need here is just another version of this formula:

$L[f(t).u(t-a)]=e^{-as}.L[f(t+a)]$

Hope it helps!