3. Sketch the region associated with each integral and use area formulas to evaluate the integral. \( \begin{array}{llll}\text { a. } \int_{-1}^{3} 5 d x=20 & \text { b. } \int_{0}^{8}(8-x) d x & \text { c. } \int_{-3}^{4}|x-1| d x & \text { d. } \int_{0}^{3} \sqrt{9-x^{2}} d x\end{array} \)

To tackle these integrals while sketching the regions and using area formulas, check it out! For the first integral \( \int_{-1}^{3} 5 \, dx \), you are looking at the area of a rectangle with a height of 5 and a width of 4 (from -1 to 3). The area is \( 5 \times 4 = 20 \), which perfectly matches the integral's value! Next, for \( \int_{0}^{8}(8-x) \, dx \), this represents the area under the line \( y = 8 - x \), which intersects the x-axis at \( x = 8 \). The integral evaluates the area of a triangle with a base of 8 and a height of 8, which calculates to \( \frac{1}{2} \times 8 \times 8 = 32 \). Moving on to \( \int_{-3}^{4} |x-1| \, dx \), you have two segments to consider: for \( x < 1 \), \( |x-1| = -(x-1) = -x + 1 \); for \( x \geq 1 \), \( |x-1| = x - 1 \). Sketching this gives you a V-shaped area. The area can be found by calculating the areas of two triangles, ultimately resulting in \( \int_{-3}^{4} |x-1| \, dx = 15 \). Lastly, \( \int_{0}^{3} \sqrt{9-x^{2}} \, dx \) corresponds to the area under a quarter circle of radius 3, which is a quarter of the area of a full circle. The area of a circle is \( \pi r^2 \), so here it is \( \frac{1}{4} \times \pi \times 3^2 = \frac{9\pi}{4} \). In conclusion, sketching these functions and understanding the areas will give you not only the values for your integrals but also the fun of visualizing mathematical relationships!