Calculus 220, section 6.3 Definite Integrals & the Fundamental Theorem notes by Tim Pilachowski In Lecture 6.2 we found a Riemann sum (midpoint version) to approximate the area under a curve on an interval a ≤ x ≤ b, utilizing partitions and a series of rectangles with width xnab∆=−: Area under the curve () ( )()()xxfxxfxxfxxfn∆∗+∆∗+∆∗+∆∗≅ K321 () ( ) ()()[]xxfxfxfxfn∆∗+++= K321. We also related the area under the curve to the antiderivative, or integral, of a function. Now we formalize the mathematics into the definite integral from a to b: () ( ) () ( )[]()∫=∆∗+++=→∆badxxfxxfxfxfxfnxK3210lim . Note that the definite integral is no longer an approximation but is equal to the area under the curve. Example A: Find the area under the curve f(x) = x on the interval [0, 10]. Answer: 50 Example B: Find the area under the curve xy 2= on the interval 2 ≤ x ≤ 7. Answer: 33433427 − Example C: Find the area under the curve 12 −= xy on the interval 5 ≤ x ≤ 13. Answer: 398 Example D: Find the area under the curve xxeey−+= on the interval 0 ≤ x ≤ ln 8. Answer: 863In all four examples above, we made use of the Fundamental Theorem of Calculus: Given a function f(x) which is continuous on an interval a ≤ x ≤ b, and given F(x), an antiderivative of f(x), then () () () ()aFbFabxFdxxfba−==∫. Note that this theorem can make use of any antiderivative, so we’ll generally choose the easiest version, i.e. the one without the “+ C”. This works because, as in the examples above, we’d always get “+ C – C = 0” anyway. Now consider a rocket fired upward from the ground at velocity 80 feet per second. Its height as a function of time is given by the function ()ttth 80162+−= . The two times it at height = 0 are found by solving ()→−−=+−= 516801602tttt t = 0 sec. and t = 5 sec. The velocity of the rocket is given by the function ()()8032+−=′=tthtv . Example E, part 1: Interpret the graph of the height function ()ttth 80162+−= on the interval 0 ≤ x ≤ 5. The graph is pictured to the right, in a window which is [–1, 10] by [–10, 110]. The rocket begins at a height = 0 at time = 0. It reaches its maximum height of 100 feet at time = 2.5 sec. It hits the ground (height = 0) again at time = 5 sec. Over the course of the 5 seconds, the net change in height = 0. Example E, part 2: Interpret the graph of the velocity function ()()8032+−=′=tthtv on the interval 0 ≤ x ≤ 5. The graph is pictured to the right, in a window which is [–1, 10] by [–100, 100]. The rocket begins with a positive (upward) velocity = 80 at time = 0. It slows down until the velocity = 0 at time = 2.5 sec. It then has a negative (downward) velocity and gains downward velocity until it hits the ground with a velocity = –80 at time = 5 sec. A key concept here is the notion that velocity can be either positive or negative, with the sign determined by direction. Example E, part 3: Find and interpret “area under the curve” for the velocity function () ()8032+−=′= tthtv on the intervals a) 0 ≤ x ≤ 2.5, b) 2.5 ≤ x ≤ 5, and c) 0 ≤ x ≤ 5. Answers: 100; –100; 0 The important concept here is that while area between the curve and the x-axis which is above the x-axis is positive, area between the curve and the x-axis which is below the x-axis is negative. In a fashion similar to a positive and negative signs indicating the upward or downward direction of velocity, positive and negative area indicated location above or below the x-axis.Example F: For n = number of people infected in hundreds and t = number of days, the rate of spread of a flu infection is found to be tdtdn35 +=. How many people would we expect to become infected on days 9 to 16? Answers: 10,900 people Example G: A company’s sales have grown at a rate of tedtds01.02001800 −= where s is the amount of sales in dollars and t is the number of days. Determine the amount of sales for the first 100 days. Answer: exactly 200000 – 20000e dollars; approximately $145634 Example H: Blood flows through an artery fastest at the center and slowest next to the artery wall. In 1842 the French physician Poiseuille developed an equation for the rate of blood flow in an artery with radius 0.2 cm: ()299040 xxv −= where v is in centimeters per second and x = distance from the center in centimeters. (The equation will be different for an artery of different length and if blood pressure is not normal.) Compare the amount of blood flowing from the center to 0.1 cm with the amount of blood flowing between 0.1 and 0.2 cm from the center. (For the correct label, think in terms of “length times width = area”, in this case “cm per second times cm = 2cm per second” which is a cross-sectional amount of blood that flows by during a time interval.) () ()[]() ()[]sec/cm67.303300401.03301.04001.0330409904023331.002=−−−=−=−∫xxdxx () ()[]() ()[]sec/cm69.11.03301.0402.03302.0401.02.0330409904023332.01.02=−−−=−=−∫xxdxx A much larger amount of blood flows through the center than near the wall. If the artery gets constricted, then either less blood is able to flow through or the patient’s blood pressure increases dramatically, or both. The consequences may be catastrophic. In summary: If a function represents an amount, the derivative (= slope of the curve) gives us a corresponding rate of change. If a function represents a rate of change, the integral (= area under the curve) gives us a corresponding