Question

Will mark brainliest

Answer 1

a. The area on [0, 10] is that of a trapezoid with bases 5 and 15 and height 10, so

$\displaystyle \int_0^{10} f(x) \, dx = \frac{5+15}2\cdot10 = \boxed{100}$

b. By linearity of the definite integral, we have

$\displaystyle \int_0^{25} f(x)\,dx = \int_0^{10} f(x)\,dx + \int_{10}^{25} f(x)\,dx$

and the area on [10, 25] is another trapezoid with bases 15 and 7.5 and height 15, so that

$\displaystyle \int_{10}^{25} f(x)\,dx = \frac{15+7.5}2\cdot15 = 168.75$

Then the total area on [0, 25] is

$\displaystyle \int_0^{25} f(x)\,dx = \boxed{268.75}$

c. The area on [25, 35] is that of a triangle with base 10 and height 15. However, $f(x)$ on this interval, so we multiply this area by -1 to get

$\displaystyle \int_{25}^{35} f(x)\,dx = -\frac{10\cdot15}2 = \boxed{-75}$

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

$\displaystyle \int_{15}^{25} f(x)\,dx = \int_{15}^{25} f(x)\,dx + \int_{25}^{35} f(x)\,dx = \boxed{0}$

e. The plot of $|f(x)|$ 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

$\displaystyle \int_{15}^{35} |f(x)| \, dx = \int_{15}^{25} f(x)\,dx - \int_{25}^{35} f(x)\,dx = 2 \int_{15}^{25} f(x)\,dx = \boxed{150}$

f. Changing the order of the limits in the integral swaps the sign of the overall integral, so

$\displaystyle \int_{10}^0 f(x)\,dx = -\int_0^{10} f(x)\,dx = \boxed{-100}$