Exam 3 Review Lecture 25: Exam 3 Review Integration 1. Evaluate deﬁnite integrals. Substitution, ﬁrst fundamental theorem of calculus (FTC 1), (and hints?) 2. FTC 2: d dx � If F (x) = x � f (t) dt = f (t) a x f (t) dt, ﬁnd the graph of F , estimate F , and change variables. a 3. Riemann sums; trapezoidal and Simpson’s rules. 4. Areas, volumes. 5. Other cumulative sums: average value, probability, work, etc. There are two types of volume problems: 1. solids of revolution 2. other (do by slices) In these problems, there will be something you can draw in 2D, to be able to see what’s going on in that one plane. In solid of revolution problems, the solid is formed by revolution around the x-axis or the y-axis. You will have to decide how to chop up the solid: into shells or disks. Put another way, you must decide whether to integrate with dx or dy. After making that choice, the rest of the procedure is systematically determined. For example, consider a shape rotated around the y-axis. • Shells: height y2 − y1 , circumference 2πx, thickness dx • Disks (washers): area πx2 (or πx22 − πx21 ), thickness dy; integrate dy. Work Work = Force · Distance We need to use an integral if the force is variable. 1 Exam 3 Review Example 1: Pendulum. See Figure 1 Consider a pendulum of length L, with mass m at angle θ. The vertical force of gravity is mg (g = gravitational coeﬃcient on Earth’s surface) L θ mass m mg Figure 1: Pendulum. In Figure 2, we ﬁnd the component of gravitational force acting along the pendulum’s path F = mg sin θ. θ mg θ Figure 2: F = mg sin θ (force tangent to path of motion). 2 Exam 3 Review Is it possible to build a perpetual motion machine? Let’s think about a simple pendulum, and how much work gravity performs in pulling the pendulum from θ0 to the bottom of the pendulum’s arc. Notice that F varies. That’s why we have to use an integral for this problem. θ0 � θ0 � (Force) · (Distance) = W = 0 (mg sin θ)(L dθ) 0 �θ0 � W = −Lmg cos θ� = −Lmg(cos θ0 − 1) = mg [L(1 − cos θ0 )] 0 In Figure 3, we see that the work performed by gravity moving the pendulum down a distance L(1 − cos θ) is the same as if it went straight down. θ L L(1-cosθ) Figure 3: Eﬀect of gravity on a pendulum. In other words, the amount of work required depends only on how far down the pendulum goes. It doesn’t matter what path it takes to get there. So, there’s no free (energy) lunch, no perpetual motion machine. 3