1) The Laplace transform of a function f(t) is defined to
be
F(s) = integral_0^infty f(t) e^{-st) dt
You have
f(t) = 5t for t<=5
and = 25 for t> 5
2) Really formal discussion:
If you are doing Laplace transforms, then you likely
have encountered the Heaviside step function
H(t - t_0 ) = 1 for t> t_0 and 0 for t < t_0
so your function can be rewritten as
f(t) = 25 H( t - 5 ) + 5t [ 1 - H(t - 5) ]
Here the first term turns on at t=5 and has a value
of 25. The second term is nonzero only up to where t=5,
then it becomes zero.
3) The point of writing it this way is that we can use the
linearity of the integration process now to write our
original integral as the sum of two integrals
F(s) = integral_0^5 5t e^{-st } dt
+ integral_5^{infinity} 25 e^{ -st } dt
Both of these integrals are elementary integrals.
The first may be solved by integration by parts, and
the second is just an integral of an exponential.
Note that it is very important that the integrand doesn't
blow up, so in each case, it is important to point out
that the first integral exists for all finite values of s,
but that the second integral only exists for s>0. Otherwise
the integrand becomes infinite as t-> infinity.
4) Note that some people will just write the two integrals
by inspection, without writing the Heaviside functions, because they tacitly know that you can paste
together piecewise defined functions to make a function
over the full domain. However, what I have
written here is what they are really relying on.