a) The first integral corresponds to the area under y = f(x) on the interval [0, 3], which is a right triangle with base 3 and height 5, hence the integral is
[tex]\displaystyle \int_0^3 f(x) \, dx = \frac12 \times 3 \times 5 = \boxed{\frac{15}2}[/tex]
b) The integral is zero since the areas under the curve over [3, 4] and [4, 5] are equal but opposite in sign. In other words, on the interval [3, 5], f(x) is symmetric and odd about x = 4, so
[tex]\displaystyle \int_3^5 f(x) \, dx = \int_3^4 f(x) \, dx + \int_4^5 f(x) \, dx = \int_3^4 f(x) \, dx - \int_3^4 f(x) \, dx = \boxed{0}[/tex]
c) The integral over [5, 9] is the negative of the area of a rectangle with length 9 - 5 = 4 and height 5, so
[tex]\displaystyle \int_5^9 f(x) \, dx = -4\times5 = -20[/tex]
Then by linearity, we have
[tex]\displaystyle \int_0^9 f(x) \, dx = \left\{\int_0^3 + \int_3^5 + \int_5^9\right\} f(x) \, dx = \frac{15}2 + 0 - 20 = \boxed{-\frac{25}2}[/tex]
