1. Ch. 8 – Applications of Definite Integrals 8.1 – Integral as Net Change

2. Recall: Given a velocity function v(t)… –The position function s(t) is the antiderivative of v(t) Antiderivative means integral, which means area underneath curve of v(t) –The acceleration function a(t) is the derivative of v(t) Derivative means slope of the curve of v(t) Ex: A particle is moving along the x-axis with a velocity of: The initial position of the particle is s(0) = 9. –Find the particle’s position at t = 1. First, find an expression for s(t) using the initial position given. Second, evaluate for s(1).

3. Ex: A particle is moving along the x-axis with a velocity of: The initial position of the particle is s(0) = 9. –Find the particle’s displacement from t = 3 to t = 7. Recall that displacement is stop minus start! –Find the total distance traveled by the particle over [0, 5]. Recall that total distance is the absolute value of total displacement, since we count every displacement (positive or negative) as distance To do this by hand, we would need to find the total positive area over [0, 5] and add it to the total negative area over [0, 5]. We’ll use the calculator for this one, though:

4. Ex: A particle is moving along the x-axis with a velocity of m/s. The initial position of the particle is s(0) = 2 m. –When is the particle moving to the right in the first 5 seconds? “moving to the right” = velocity is positive We must find x-intercepts of v(t) and make a sign chart to see where v(t) > 0. Since t is on [0, 5], then Now make a sign chart for v(t): –Answer: v(t) is moving to the right on [0, π/2] and [3π/2, 5] on the interval [0, 5]. ++ -

5. Ex: Below is a graph of the velocity, v(t), of a particle moving along the x-axis. The particle is at x = 2 when t = 0. Using the graph, find… –…the location of the particle at t = 10. Position  antiderivative of v(t)  integral (area) from t = 0 to t = 10. Find areas of each section! 3 + 4.5 – 12 = -4.5 Since we started at x = 2, we end at 2 + (-4.5) = -2.5 –…the total distance traveled by the particle over [0, 10]. Distance = positive sum of areas of each section 3 + 12 + 4.5 = 19.5 3 12 4.5