To find the integral we need to divide it according to the intervals that define the piece-wise function, then we have:
[tex]\int_0^2f(x)dx=\int_0^1f(x)dx+\int_1^2f(x)dx[/tex]
Now that we have divide the integral we plug the expressions that define it in the intervals:
[tex]\begin{gathered} \int_0^2f(x)dx=\int_0^16x^4dx+\int_1^23x^5dx \\ =(\frac{6}{5}x^5)\vert_0^1+(\frac{3}{6}x^6)\vert_1^2 \\ =(\frac{6}{5}(1)^5-\frac{6}{5}(0)^5)+(\frac{3}{6}(2)^6-\frac{3}{6}(1)^6) \\ =\frac{6}{5}+32-\frac{1}{2} \\ =\frac{327}{10} \end{gathered}[/tex]
Therefore, the integral is 327/10
