a. The area on [0, 10] is that of a trapezoid with bases 5 and 15 and height 10, so
[tex]\displaystyle \int_0^{10} f(x) \, dx = \frac{5+15}2\cdot10 = \boxed{100}[/tex]
b. By linearity of the definite integral, we have
[tex]\displaystyle \int_0^{25} f(x)\,dx = \int_0^{10} f(x)\,dx + \int_{10}^{25} f(x)\,dx[/tex]
and the area on [10, 25] is another trapezoid with bases 15 and 7.5 and height 15, so that
[tex]\displaystyle \int_{10}^{25} f(x)\,dx = \frac{15+7.5}2\cdot15 = 168.75[/tex]
Then the total area on [0, 25] is
[tex]\displaystyle \int_0^{25} f(x)\,dx = \boxed{268.75}[/tex]
c. The area on [25, 35] is that of a triangle with base 10 and height 15. However, [tex]f(x)<0[/tex] on this interval, so we multiply this area by -1 to get
[tex]\displaystyle \int_{25}^{35} f(x)\,dx = -\frac{10\cdot15}2 = \boxed{-75}[/tex]
d. The area on [15, 25] is the same as the area on [25, 35] because it's another triangle with the same dimensions. But the area on [15, 25] lies above the horizontal axis, so
[tex]\displaystyle \int_{15}^{25} f(x)\,dx = \int_{15}^{25} f(x)\,dx + \int_{25}^{35} f(x)\,dx = \boxed{0}[/tex]
e. The plot of [tex]|f(x)|[/tex] lies above the horizontal axis. We know the area on [15, 25] is the same as the area on [25, 35], but now both areas are positive, so
[tex]\displaystyle \int_{15}^{35} |f(x)| \, dx = \int_{15}^{25} f(x)\,dx - \int_{25}^{35} f(x)\,dx = 2 \int_{15}^{25} f(x)\,dx = \boxed{150}[/tex]
f. Changing the order of the limits in the integral swaps the sign of the overall integral, so
[tex]\displaystyle \int_{10}^0 f(x)\,dx = -\int_0^{10} f(x)\,dx = \boxed{-100}[/tex]
