1. 4.7 Improper Integrals 1 For integration, we have assumed 1. limits of integration are finite, 2. integrand f(x) is bounded. The violation of either of these two conditions makes the integral improper. Improper Integral Type I: the integrand has an infinite discontinuity in the interval [a,b]: NOT covered by the class!

2. 2 Improper Integral Type II: one of the limits of integration is infinite. Example: Evaluate the given integral: Solution: We substitute the infinity by a variable, evaluate the integral, and then take a limit as the variable approaches the infinity.

3. 3 Exercise: Find the area of the region bounded by, the coordinate axes; area in the first quadrant.

4. 4.8 The Constant of Integration and Applications of the Indefinite Integral 4 Agenda: Usually in physics and other disciplines, the acceleration is given, and the trajectory (distance as a function of time) is to be determined by integration. We start with a simpler, geometric problem: Example: Find the function satisfying the conditions Solution: Integrate the equality first: This form defines a family of curves because the constant C is not fixed. This family includes the curves that differ from each other by a vertical shift. To define a single curve, we use the extra condition.

5. 5 Example (cntd): We plug the numbers y=3 and x=-2 to get the equation for C: Hence, Finally, we plug this value of C into the function:

6. The problem of finding the trajectory of a body under the influence of a force We can now rewrite the laws of Section 2.6: in the integral form: To describe the vertical motion of an object under the influence of gravity, we assume: 1. a≡g=-10m/s 2 (the acceleration is given as a constant number!) 2. s=0 on the ground. 3. t=0 at the instant when the motion begins. 4. A distance from the ground to a point above the ground is positive. 5. An object moving in the upward direction has a positive velocity. 6. An object moving downward has a negative velocity. 6

7. 7 Example: A ball is thrown upward with a velocity of 20 m/s. a. How high does it rise? b. How long does it stay in the air? Solution: Since the ball is thrown in the upward direction, v(0)=v 0 =+20m/s. This is the initial velocity. But gravity accelerates the ball in the downward direction, and, consequently, g=–10m/s 2. We use the initial velocity to determine the constant of integration C: Thus, where K is a constant of integration. Since the initial position of the ball is on the ground, where s=0, K=_____, and

8. 8 Example (cntd) a. How high does the ball rise? At the highest point, the velocity is zero (if we solve this rigorously, we need to examine the distance s on maxima as we did another time, and the step we take now is just a shortcut; think why) This is the instant when the maximum is attained, and the height at this instant is b. How long does it stay in the air? The ball hits the ground when s=0. or The roots of this equation are t=0 and t=4. The first one is when the motion begins, the ball returns to the ground at t=4. So, the ball stays in the air for 4 seconds.

9. 9 Homework: Section 4.7: 1,3,11,15,19. Section 4.8: 1,5,7,9,13,21,23,25,27,29.