1. **State the problem:** We need to find the value of the definite integral $$\int_{-8}^{4} f(x) \, dx$$ where the function $f$ consists of three parts: a line segment from $(-8,5)$ to $(-5,-5)$, an upper semicircle from $(-5,-5)$ to $(5,-5)$ with top at $(0,0)$, and a line segment from $(5,-5)$ to $(7,-3)$. Since the integral is from $-8$ to $4$, we only consider the first line segment and part of the semicircle. 2. **Break the integral into parts:** $$\int_{-8}^{4} f(x) \, dx = \int_{-8}^{-5} f(x) \, dx + \int_{-5}^{4} f(x) \, dx$$ 3. **First part: line segment from $(-8,5)$ to $(-5,-5)$** - The slope is $$m = \frac{-5 - 5}{-5 - (-8)} = \frac{-10}{3} = -\frac{10}{3}$$ - Equation of the line: $$y - 5 = -\frac{10}{3}(x + 8)$$ $$y = -\frac{10}{3}x - \frac{80}{3} + 5 = -\frac{10}{3}x - \frac{80}{3} + \frac{15}{3} = -\frac{10}{3}x - \frac{65}{3}$$ 4. **Integral over $[-8,-5]$:** $$\int_{-8}^{-5} \left(-\frac{10}{3}x - \frac{65}{3}\right) dx = \left[-\frac{10}{3} \cdot \frac{x^2}{2} - \frac{65}{3}x \right]_{-8}^{-5} = \left[-\frac{5}{3}x^2 - \frac{65}{3}x \right]_{-8}^{-5}$$ Calculate at bounds: - At $x=-5$: $$-\frac{5}{3}(-5)^2 - \frac{65}{3}(-5) = -\frac{5}{3} \cdot 25 + \frac{325}{3} = -\frac{125}{3} + \frac{325}{3} = \frac{200}{3}$$ - At $x=-8$: $$-\frac{5}{3}(-8)^2 - \frac{65}{3}(-8) = -\frac{5}{3} \cdot 64 + \frac{520}{3} = -\frac{320}{3} + \frac{520}{3} = \frac{200}{3}$$ So the integral over $[-8,-5]$ is: $$\frac{200}{3} - \frac{200}{3} = 0$$ 5. **Second part: semicircle from $x=-5$ to $x=5$ with center at $(0,-5)$ and radius $5$** - Equation of the circle: $$ (x - 0)^2 + (y + 5)^2 = 5^2 = 25 $$ - Solve for $y$ (upper semicircle): $$ y = -5 + \sqrt{25 - x^2} $$ 6. **Integral over $[-5,4]$:** $$ \int_{-5}^{4} \left(-5 + \sqrt{25 - x^2}\right) dx = \int_{-5}^{4} -5 \, dx + \int_{-5}^{4} \sqrt{25 - x^2} \, dx $$ Calculate each separately: - $$ \int_{-5}^{4} -5 \, dx = -5 (4 - (-5)) = -5 \times 9 = -45 $$ - $$ \int_{-5}^{4} \sqrt{25 - x^2} \, dx $$ is the area under the upper semicircle from $x=-5$ to $x=4$. 7. **Integral of semicircle segment:** The integral $$\int \sqrt{r^2 - x^2} \, dx$$ has the formula: $$ \frac{x}{2} \sqrt{r^2 - x^2} + \frac{r^2}{2} \arcsin\left(\frac{x}{r}\right) + C $$ Apply with $r=5$: $$ \int_{-5}^{4} \sqrt{25 - x^2} \, dx = \left[ \frac{x}{2} \sqrt{25 - x^2} + \frac{25}{2} \arcsin\left(\frac{x}{5}\right) \right]_{-5}^{4} $$ Calculate at bounds: - At $x=4$: $$ \frac{4}{2} \sqrt{25 - 16} + \frac{25}{2} \arcsin\left(\frac{4}{5}\right) = 2 \times 3 + \frac{25}{2} \arcsin\left(0.8\right) = 6 + \frac{25}{2} \arcsin(0.8) $$ - At $x=-5$: $$ \frac{-5}{2} \sqrt{25 - 25} + \frac{25}{2} \arcsin\left(-1\right) = 0 + \frac{25}{2} \times (-\frac{\pi}{2}) = -\frac{25\pi}{4} $$ So the integral is: $$ \left(6 + \frac{25}{2} \arcsin(0.8)\right) - \left(-\frac{25\pi}{4}\right) = 6 + \frac{25}{2} \arcsin(0.8) + \frac{25\pi}{4} $$ 8. **Combine the parts:** $$ \int_{-8}^{4} f(x) \, dx = 0 + \left(-45 + 6 + \frac{25}{2} \arcsin(0.8) + \frac{25\pi}{4}\right) = -39 + \frac{25}{2} \arcsin(0.8) + \frac{25\pi}{4} $$ 9. **Final answer:** $$ \boxed{ -39 + \frac{25}{2} \arcsin\left(\frac{4}{5}\right) + \frac{25\pi}{4} } $$ This is the exact value of the integral in simplest form.