F (x) is an antiderivative of f(x) if F (x) = f(x). Lets find an antiderivative of f(x) = x. We know that d. Any ideas?

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1 Math 24 - Calculus for Management and Social Science Antiderivatives and the Indefinite Integral: Notes So far we have studied the slope of a curve at a point and its applications. This is one of the fundamental problems in calculus. The other fundamental problem in calculus is to find the area of a region under a curve. It turns out these problems are intimately connected. So far we have developed rules for finding the derivative of a function. Now it is necessary to do the reverse. Given the derivative of a function, determine the function. With this in mind we have the following definition: F () is an antiderivative of f() if F () = f(). Lets find an antiderivative of f() =. We know that d d (2 ) = 2, which is almost the same as ecept for the 2. Any ideas? How about F () = 2 2? Well, F () =. So F () = 2 2 is an antiderivative of. How about F () = 2 2 +? Of course this one is also an antiderivative of. In fact, for any constant C, F () = C is an antiderivative of. If F () and G() are both antiderivatives of f() then F () G() = C, where C is a constant. The graphs differ by a vertical shift. The set of all antiderivatives of f() can be described by F () + C, where C is a constant and F () is a particular antiderivative. That motivates our net definition The indefinite integral of f() is the set of all antiderivatives, denoted in the following way: f()d = F () + C. is the integral sign. C is a constant and F () = f().

2 Consider the following derivative for n. ( ) d d n + n+ = n + d ( ) n+ = d n + (n + )n = n. So F () = n+ n+ is an antiderivative of f() = n for n. Thus, for n, and arbitrary constant C, n d = n + n+ + C. Here are a few eamples: 3 d = C. 2 d = + C. d = C. What about n =? That is, what is an antiderivative of f() = =? We know that F () = f() for F () = ln. But since the domain of F () = ln does not match the domain of f() = we must use G() = ln as the antiderivative since G () = f() and the domain matches. Hence, d = ln + C, where 0. Consider the following derivative for k 0. Hence, for a constant k 0, ( ) d d k ek = k d ( ) e k = d k kek = e k. e k d = k ek + C.

3 We also have the following nice rules, just as for derivatives, [f() ± g()]d = f()d ± g()d. kf()d = k f()d, where k is a constant. Eample: ( ) d = 6 d d + 3 d = = ln + C = ln + C. Find a function F () that satisfies F () = 4e.8 and F (0) = 0. Lets begin with the indefinite integral: F () = 4e.8 d = 4 e.8 d = 4 (.8) e.8 + C = 5e.8 + C. To determine C use that 0 = F (0) = 5 + C. Hence C = 5 and F () = 5e The Definite Integral: For a continuous nonnegative function on the interval [a, b], we refer to the area under the curve as the area of the region bounded by the the curve and the -ais over [a, b]. For eample, by the area under y = 2 on [ 3, 5], we mean the following shaded area:

4 We can approimate the area under the curve using rectangles. For eample, we can divide [ 3, 5] into 8 equal subintervals of width. Then 8 rectangles with height given by the function s value at the left endpoints of the subintervals can be used to approimate the area under the curve. We could use any points within the subintervals, such as midpoints or right endpoints. s u m A = 44. Of course, the more rectangles used the better. Below gives us the estimate with 48 rectangles with height given by the function s value at left endpoints. s u m

5 In general, for y = f(), when you divide [a, b] into n equal subintervals, each subinterval has width given by = b a n. Let, 2,..., n be the sample points in increasing order, one from each of the n subintervals. On the previous page, these were taken to be left endpoints. Then the area of the first rectangle is A =(height)(width)= f( ). And the area of the second rectangle is A 2 =(height)(width)= f( 2 ), and so on it goes. So the area of the rectangles, called a Riemann sum, is given by A = (f( ) + f( 2 ) + + f( n )). Lets use a Riemann sum with n = 6 and midpoints as sample points, to estimate the area under y = 3 on [, 3]. First of all, = 3 6 = 2. Our sample points are the subinterval midpoints: 2, 4, 6, 8, 0, 2. So our Riemann sum is ( ) 2 = ( ) 2 = = Of course the more rectangles, the better the sum comes to approimating the area. Now we let the function be negative, zero or positive. The limit of the Riemann sum as the number of rectangles, n, grows large (which means the widths are approaching zero) is the definite integral and measures the net area, which is the area above the -ais minus the area below the -ais. If f() is continuous on [a, b], then the definite integral is given by b a f()d = lim n (f( ) + f( 2 ) + + f( n )). The Fundamental Theorem of Calculus stated below is the connection between definite integrals and antiderivatives: b a f()d = F (b) F (a), where F () is an antiderivative of f().

6 Lets compute which we approimated on the last page. 3 3 d, We know that F () = 4 4 is an antiderivative of 3, so 3 3 d = F (3) F () = = 740. So our estimate on the last page is an underestimate, but not too far off! Lets solve for the area depicted on page three. F () = 3 3 is an antiderivative of d = F (5) F ( 3) = 25 3 ( 27) 3 = , which is close to our estimate using n = 48.