REVIEW NOTES FOR FIRST MATH 1B EXAM PETE L CLARK ADAPTED BY RG 1 Riemann integrals You should know the definition of the definite integral of a continuous function f on a closed interval a b as a limit of Riemann sums a Riemann sum is obtained by dividing the interval a b into n subintervals of equal length x b a n by selecting a point xi on the subinterval xi xi 1 and approximating the value of the function on the entire subinterval by the value f x i thus we are approximating the area on xi xi 1 by xf x i b a n f xi Pn 1 the entire Riemann sum is then i 0 f x i x Then if f is continuous no matter which sample points x i are Rb Pn 1 chosen these Riemann sums will converge to a common value in the limit limn i 0 f x i x a f x dx This value is by definition the definite integral The Riemann sum can be taken for a function which is both positive and negative signed area i e the positive area the negative area Rb a f x dx represents the The fundamental theorem of calculus gives us a powerful method for computing these definite integrals which would be anywhere from needlessly onerous to nigh impossible to compute directly from the definition if F x is Rb an antiderivative of f x that is a function such that F 0 f then we know a f x dx F x ba F b F a Note that an antiderivative of a continuous function on a closed interval a b is well determined up to an additive constant if F1 F2 are two functions such that F10 f F20 then F1 F2 0 F10 F20 f f 0 so F1 F2 has zero derivative on a b hence is constant It is important to remember this indeterminacy e g Question A particle is moving along a line with the velocity function v t sin t If its initial position is x 0 0 what is its position at any time t R Solution By definition v t x0 t so that x t v t cos t C What is C It is such that 0 x 0 cos 0 C 1 C i e C 1 and the position function is x t 1 cos t Notice that if we forgot to include the constant C our answer would have been wrong Another warning many students become so enamored of the fundamental theorem of calculus that they forget all about the definition of the definite integral as the limit of Riemann sums We instructors like to force you to remember the geometric definition of the definite integral by giving problems where the definite integral can be easily computed when finding the antiderivative is difficult or impossible Our favorite example of this is the following Fact If f x is an odd function f x f x or otherwise put the graph of f is symmetric about the origin Ra R0 Ra R 2 then a f x dx 0 for a any finite number since a f x dx 0 f x dx For example sin x ex dx would be a real pain to find the antiderivative but the definite integral is 0 Rb Similarly you should have some idea of how to use Riemann sums to get an approximate value for a f x dx although we would not ask you to numerically evaluate a Riemann sum on an exam we might ask you to argue geometrically about the value of a definite integral using Riemann sums for example Question Give a method for calculating R1 0 2 ex dx to within an accuracy of 000001 2 Solution First recall that ex is our basic example of a function whose antiderivative we cannot write down 1 2 PETE L CLARK ADAPTED BY RG so we must proceed numerically The idea of course is to divide into a bunch of subintervals and take a Riemann 2 sum but how will we know how far we are We exploit the fact that f x ex is an increasing function on 2 0 1 check f 0 x 2xex 0 on this interval so that the left endpoint sums will be lower sums and the right endpoint sums will be upper sums draw a picture Here x n1 and xi ni so the left endpoint sum is Pn 1 i 2 i Pn 1 i 1 2 i n n i 0 e i 0 e n whereas the right endpoint sum is n We know the first quantity is always less than the second so we get a calculational device TI 85 or mathematical software package to spit out these sums for increasing n until we find a value of n for which these two quantities differ by less than 000001 2 Geometric applications of integration area volume arclength 2 1 Area between curves If f x and g x are two functions defined on a b we can roughly speaking integrate their difference to find the area between them However we must be careful about which function is Rb on top technically we define the area between f and g to be a f x g x dx but what this really means is that we must break up the integral into regions where f x g x has constant sign always positive or always negative When f x g x is positive f x g x f x g x so we integrate f x g x But when f x g x is negative f x g x g x f x so we must integrate the latter Example Find the area enclosed by y sin x and y 0 on 0 2 Solution The answer is R 2 0 sin x dx R 0 sin xdx R 2 sin xdx because sin x 0 for x 2 2 2 Volumes by slicing If we want to find the volume of a three dimensional region whose cross sectional areas are known to us then we can simply integrate the cross sectional area to get the volume Z b V A x dx a 2 3 Volumes of revolution If we are given a volume of revolution i e a solid obtained by revolving a region bounded by curves about an axis parallel to the x or y axis we have three methods for finding the volume of this region the methods of disks washers and cylindrical shells Disks and Washers really just amount to special cases of slicing like you would slice a loaf of bread If we are revolving around an axis parallel to the x axis and we wish to integrate with respect to x then we use the method of washers or disks Similarly if we are revolving around an axis parallel to the y axis and we wish to integrate with respect to y then we use washers or disks A disk is just a washer whose inner radius is …