1. **Problem statement:** We need to find the range of values of $a$, with $0 \leq a \leq 5$, for which the integral $$\int_0^a f(x) \, dx$$ is positive. 2. **Understanding the graph:** The function $f(x)$ is defined on $[0,5]$ and consists of three line segments: - From $x=0$ to $x=1$, $f(x) = 2$ (horizontal line). - From $x=1$ to $x=2$, $f(x)$ decreases linearly from 2 to -2. - From $x=2$ to $x=5$, $f(x) = -2$ (horizontal line). 3. **Integral interpretation:** The integral $$\int_0^a f(x) \, dx$$ represents the net area between the curve and the $x$-axis from 0 to $a$. Positive area is above the $x$-axis, negative area is below. 4. **Calculate areas step-by-step:** - For $0 \leq a \leq 1$: $$\int_0^a f(x) \, dx = \int_0^a 2 \, dx = 2a$$ This is positive for all $a > 0$. - For $1 < a \leq 2$: The integral is the sum of the area from 0 to 1 plus the area from 1 to $a$. Area from 0 to 1: $$\int_0^1 2 \, dx = 2 \times 1 = 2$$ Area from 1 to $a$: Since $f(x)$ decreases linearly from 2 to -2, slope $m = \frac{-2 - 2}{2 - 1} = -4$. Equation of line segment: $$f(x) = 2 - 4(x - 1) = 6 - 4x$$ Integral from 1 to $a$: $$\int_1^a (6 - 4x) \, dx = \left[6x - 2x^2\right]_1^a = (6a - 2a^2) - (6 - 2) = 6a - 2a^2 - 4$$ Total integral from 0 to $a$: $$2 + (6a - 2a^2 - 4) = 6a - 2a^2 - 2$$ - For $2 < a \leq 5$: The integral is the sum of the integral from 0 to 2 plus the integral from 2 to $a$. Calculate integral from 0 to 2: $$\int_0^1 2 \, dx + \int_1^2 (6 - 4x) \, dx = 2 + \left[6x - 2x^2\right]_1^2 - (6 - 2) = 2 + (12 - 8) - 4 = 2 + 4 - 4 = 2$$ From 2 to $a$, $f(x) = -2$: $$\int_2^a -2 \, dx = -2(a - 2) = -2a + 4$$ Total integral from 0 to $a$: $$2 + (-2a + 4) = 6 - 2a$$ 5. **Find where the integral is positive:** - For $0 \leq a \leq 1$: $$2a > 0 \implies a > 0$$ - For $1 < a \leq 2$: $$6a - 2a^2 - 2 > 0$$ Divide by 2: $$3a - a^2 - 1 > 0$$ Rewrite: $$-a^2 + 3a - 1 > 0$$ Multiply both sides by $-1$ (flip inequality): $$a^2 - 3a + 1 < 0$$ Solve quadratic inequality: $$a = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$$ Approximate roots: $$a \approx 0.38 \text{ and } 2.62$$ Since $a$ is in $(1,2]$, the inequality holds for $a$ between 0.38 and 2.62, intersected with $(1,2]$ gives $1 < a < 2$. - For $2 < a \leq 5$: $$6 - 2a > 0 \implies 2a < 6 \implies a < 3$$ 6. **Combine all intervals:** - From $0 < a \leq 1$, integral positive. - From $1 < a < 2$, integral positive. - From $2 < a < 3$, integral positive. - For $a \geq 3$, integral is zero or negative. 7. **Final answer:** $$\boxed{0 < a < 3}$$ The integral $$\int_0^a f(x) \, dx$$ is positive for all $a$ in the interval $(0,3)$.