1. **Problem Statement:** Evaluate the following definite integrals of the function $f(x)$ whose graph consists of a semicircle below the x-axis from $x=0$ to $x=2$ and two line segments from $x=2$ to $x=4$. 2. **Given:** - From $0$ to $2$, $f(x)$ is a semicircle below the x-axis with diameter on the $x$-axis from $(0,0)$ to $(2,0)$ and minimum $y=-1$. - From $2$ to $3$, $f(x)$ is a line segment from $(2,0)$ to $(3,2)$. - From $3$ to $4$, $f(x)$ is a line segment from $(3,2)$ to $(4,0)$. 3. **Formulas and rules:** - Area of a semicircle: $$\text{Area} = \frac{1}{2} \pi r^2$$ - Area of a triangle: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ - The definite integral of $f(x)$ over an interval equals the net area between the graph and the $x$-axis. - For integrals involving $x f(x)$, use geometric interpretation or break into sums if possible. 4. **Step-by-step solutions:** **(a) $\int_0^2 f(x) \, dx$** - The graph is a semicircle below the $x$-axis with radius $r=1$ (since diameter is 2). - Area of full circle: $\pi r^2 = \pi \times 1^2 = \pi$. - Area of semicircle: $\frac{1}{2} \pi = \frac{\pi}{2}$. - Since the semicircle is below the $x$-axis, integral is negative: $$\int_0^2 f(x) \, dx = -\frac{\pi}{2}$$ **(b) $\int_2^4 f(x) \, dx$** - From 2 to 3: triangle with base 1 and height 2, area = $\frac{1}{2} \times 1 \times 2 = 1$. - From 3 to 4: triangle with base 1 and height 2, area = $\frac{1}{2} \times 1 \times 2 = 1$. - Both triangles are above the $x$-axis, so integral is sum of areas: $$\int_2^4 f(x) \, dx = 1 + 1 = 2$$ **(c) $\int_0^3 x f(x) \, dx$** - Break integral into two parts: $$\int_0^2 x f(x) \, dx + \int_2^3 x f(x) \, dx$$ - For $0 \leq x \leq 2$, $f(x)$ is semicircle below $x$-axis. Use symmetry and geometry: The semicircle is centered at $x=1$ with radius 1. The function $f(x)$ is negative semicircle: $f(x) = -\sqrt{1-(x-1)^2}$. - Compute $\int_0^2 x f(x) \, dx$: Use substitution $u = x-1$, so $x = u+1$, $u \in [-1,1]$. Then: $$\int_0^2 x f(x) \, dx = \int_{-1}^1 (u+1)(-\sqrt{1-u^2}) \, du = -\int_{-1}^1 (u+1) \sqrt{1-u^2} \, du$$ - Split integral: $$-\left( \int_{-1}^1 u \sqrt{1-u^2} \, du + \int_{-1}^1 \sqrt{1-u^2} \, du \right)$$ - The function $u \sqrt{1-u^2}$ is odd, so integral over $[-1,1]$ is 0. - The integral $\int_{-1}^1 \sqrt{1-u^2} \, du$ is area of a semicircle of radius 1, which is $\frac{\pi}{2}$. - So: $$\int_0^2 x f(x) \, dx = -\frac{\pi}{2}$$ - For $2 \leq x \leq 3$, $f(x)$ is line from $(2,0)$ to $(3,2)$. Equation of line: $$f(x) = 2(x-2)$$ - Compute: $$\int_2^3 x f(x) \, dx = \int_2^3 x \cdot 2(x-2) \, dx = 2 \int_2^3 x(x-2) \, dx = 2 \int_2^3 (x^2 - 2x) \, dx$$ - Calculate integral: $$2 \left[ \frac{x^3}{3} - x^2 \right]_2^3 = 2 \left( \left(\frac{27}{3} - 9\right) - \left(\frac{8}{3} - 4\right) \right) = 2 \left( (9 - 9) - (\frac{8}{3} - 4) \right) = 2 \left( 0 - (\frac{8}{3} - 4) \right) = 2 \left( 4 - \frac{8}{3} \right) = 2 \times \frac{12 - 8}{3} = 2 \times \frac{4}{3} = \frac{8}{3}$$ - Sum parts: $$\int_0^3 x f(x) \, dx = -\frac{\pi}{2} + \frac{8}{3}$$ **(d) $\int_0^4 f(x) \, dx$** - Sum integrals from (a) and (b): $$\int_0^4 f(x) \, dx = \int_0^2 f(x) \, dx + \int_2^4 f(x) \, dx = -\frac{\pi}{2} + 2$$ **(e) $\int_0^4 [f(x) + 2] \, dx$** - Use linearity: $$\int_0^4 f(x) \, dx + \int_0^4 2 \, dx = \left(-\frac{\pi}{2} + 2\right) + 2 \times 4 = -\frac{\pi}{2} + 2 + 8 = -\frac{\pi}{2} + 10$$ 5. **Final answers:** - (a) $\boxed{-\frac{\pi}{2}}$ - (b) $\boxed{2}$ - (c) $\boxed{-\frac{\pi}{2} + \frac{8}{3}}$ - (d) $\boxed{-\frac{\pi}{2} + 2}$ - (e) $\boxed{-\frac{\pi}{2} + 10}$