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 b a x a x b utilizing partitions and a series of rectangles with width n Area under the curve f x1 x f x 2 x f x3 x K f x n x f x1 f x 2 f x3 K f x n x 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 lim x 0 f x1 f x2 f x3 K f xn x a f x dx b 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 y 2 x on the interval 2 x 7 Answer 4 7 3 4 2 3 3 3 Example C Find the area under the curve y 2 x 1 on the interval 5 x 13 Answer 98 3 Example D Find the area under the curve y e x e Answer 63 8 x on the interval 0 x ln 8 In 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 b b then f x dx F x F b F a a a 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 h t 16t 2 80t The two times it at height 0 are found by solving 0 16t 2 80t 16t t 5 t 0 sec and t 5 sec The velocity of the rocket is given by the function v t h t 32t 80 Example E part 1 Interpret the graph of the height function h t 16t 2 80t 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 v t h t 32t 80 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 v t h t 32t 80 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 dn infection is found to be 5 3 t How many people would we expect to become infected on days 9 to 16 dt Answers 10 900 people ds 1800 200e 0 01t where s is the amount of sales in dt 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 G A company s sales have grown at a rate of 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 v x 40 990 x 2 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 cm 2 per second which is a cross sectional amount of blood that flows by during a time interval 0 1 0 0 2 0 1 40 990 x 2 dx 40 x 330 x 3 40 990 x 2 dx 40 x 330 x 3 0 1 40 0 1 330 0 1 3 40 0 330 0 3 3 67 cm 2 sec 0 0 2 0 1 40 0 2 330 0 2 3 40 0 1 330 0 1 3 1 69 cm 2 sec 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 amount