Question

the figure to the right shows the area of regions bounded by the graph of f and
the x - axis
evaluate the following integral
\\( \int _ { a } ^ { c } f ( x ) d x \\)

Explanation:

Step1: Understand Integral as Area

The definite integral $\int_{a}^{c} f(x) dx$ represents the net area between the curve $y = f(x)$, the x - axis, from $x=a$ to $x = c$. Areas above the x - axis are positive, and areas below the x - axis are negative.

Step2: Identify Areas

• From $x = a$ to $x = b$: The area is below the x - axis (the brown region) with area 8, so its contribution to the integral is $- 8$.
• From $x = b$ to $x = c$: The area is above the x - axis (the green region) with area 12, so its contribution to the integral is $+ 12$. Also, from $x = 0$ to $x=a$, the area is above the x - axis with area 18, but wait, no, the integral is from $a$ to $c$. Wait, let's re - examine. Wait, the figure: from $a$ to $b$, the region is below x - axis (area 8, so integral contribution $-8$), from $b$ to $c$, the region is above x - axis (area 12, integral contribution $+12$), and also, wait, maybe I misread. Wait, the integral is $\int_{a}^{c}f(x)dx$. So we split the integral into $\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx$.

For $\int_{a}^{b}f(x)dx$: the area between $a$ and $b$ is below the x - axis, so the integral is $- 8$ (since area below x - axis is negative in definite integral).

For $\int_{b}^{c}f(x)dx$: the area between $b$ and $c$ is above the x - axis, so the integral is $+ 12$.

Wait, but also, is there a region from $0$ to $a$? No, the integral is from $a$ to $c$. So $\int_{a}^{c}f(x)dx=\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx=- 8 + 12=4$? Wait, no, maybe I made a mistake. Wait, the figure: the area from $0$ to $a$ is 18 (above x - axis), from $a$ to $b$ is 8 (below x - axis), from $b$ to $c$ is 12 (above x - axis). But the integral is from $a$ to $c$. So $\int_{a}^{c}f(x)dx=\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx$. The area below x - axis (from $a$ to $b$) gives a negative integral, and area above x - axis (from $b$ to $c$) gives a positive integral. So $\int_{a}^{b}f(x)dx=- 8$ (because it's below x - axis), $\int_{b}^{c}f(x)dx = 12$ (because it's above x - axis). Then $\int_{a}^{c}f(x)dx=-8 + 12 = 4$? Wait, no, wait, maybe the area from $a$ to $b$ is a "negative" area (since it's below x - axis) and from $b$ to $c$ is positive. So adding them: $-8+12 = 4$? Wait, but let's check again.

Wait, the definite integral $\int_{a}^{c}f(x)dx$ is the net area, which is (area above x - axis from $b$ to $c$) minus (area below x - axis from $a$ to $b$). Wait, no: if the function is below the x - axis, $f(x)$ is negative, so the integral $\int_{a}^{b}f(x)dx$ is negative (equal to - area of the region). If the function is above the x - axis, $\int_{b}^{c}f(x)dx$ is positive (equal to the area of the region). So the region from $a$ to $b$: area is 8, and since it's below x - axis, $\int_{a}^{b}f(x)dx=-8$. The region from $b$ to $c$: area is 12, above x - axis, so $\int_{b}^{c}f(x)dx = 12$. Then $\int_{a}^{c}f(x)dx=\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx=-8 + 12=4$. Wait, but what about the area from $0$ to $a$? Oh, the integral is from $a$ to $c$, so we don't care about $0$ to $a$ here.

Answer:

$\boxed{4}$