Since [tex][0,4]=[0,1]\cup(1,4][/tex], we can rewrite the integral as
[tex]\displaystyle \int_0^1f(t)\;dt + \int_1^4 f(t)\; dt[/tex]
Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:
[tex]\displaystyle \int_0^1f(t)\;dt = \int_0^11-3t^2\;dt,\quad \int_1^4 f(t)\; dt = \int_1^4 2t\; dt[/tex]
Both integrals are quite immediate: you only need to use the power rule
[tex]\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}[/tex]
to get
[tex]\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4[/tex]
Now we only need to evaluate the antiderivatives:
[tex]\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15[/tex]
So, the final answer is 15.
