45 Now we set up the following inequality E S n 24 1 5180 n 4 1 1000

# 45 now we set up the following inequality e s n 24 1

• 200
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 51 - 55 out of 200 pages.

45
Now we set up the following inequality E S n 24 · 1 5 180 n 4 1 1000 Manipulating this inequality we have n 4 400 3 . Thus we write our final answer as Let n be the smallest even integer greater than 4 q 400 3 . Using a calculator, we see that n = 4 subintervals is sufficient to guarantee that our approximation is within 1 1000 of ln 2. Note. In our final answer, we had to write “Let n be the smallest even integer . . . Unlike the previous example, we had to specify that n was an even integer because we were using Simpson’s Rule, which requires an even number of subintervals. 8.2 Improper Integrals There are two types of integrals that are described as improper . I. f ( x ) is a bounded function, but the interval of integration is unbounded. For instance, in the graph below the interval is [ a, ). II. g ( x ) is an unbounded function, but the interval of integration is bounded. For instance, in the graph below the interval is [0 , 1], but the function has a vertical asymptote at x = 0. 46
Question : Can we define the area under the graph in either scenario? I. For any c > a , R c a f ( x ) dx is defined. If R c a f ( x ) dx approaches a value as c increases without bound, then we define Z a f ( x ) dx = lim c →∞ Z c a f ( x ) dx II. Suppose we wish to evaluate R b a g ( x ) dx , g ( x ) is continuous on ( a, b ], and g ( x ) has a vertical asymptote at x = a . The R b c g ( x ) dx is defined for all a < c b . If R a c g ( x ) dx approaches a value as c approaches a from above, then we define Z b a g ( x ) dx = lim c a + Z b c g ( x ) dx In either scenario, if the limit exists as a real number then we say the integral converges . Otherwise it diverges . 8.2.1 Type I Improper Integrals We will begin by investigating integrals in which our function is bounded, but we are integrating over an unbounded interval. Example 8.3. Determine whether Z 1 1 x 2 dx converges or diverges. If it con- verges, to what value does it converge? We wish to find the area of the region shaded in the graph below. Is this an infinite amount of area? Or is the amount of area under the curve bounded by a value as we let our upper bound increase? 47
To determine whether or not this converges, we convert this integral to a limit, as below. Z 1 1 x 2 dx = lim c →∞ Z c 1 1 x 2 dx = lim c →∞ - 1 x c 1 = lim c →∞ - 1 c + 1 = 1 Thus the integral converges to 1 . Note. No matter how large we let c be, the amount of area under the curve on [1 , c ] is less than 1, but getting closer and closer to 1 as we let c be larger. This is why we can conclude that there’s not an infinite amount of area under the curve despite the interval being [1 , ). Example 8.4. Determine whether Z 1 1 x dx converges or diverges. If it con- verges, to what value does it converge? We approach this problem in the exact same way we approached the previous problem. Z 1 1 x dx = lim c →∞ Z c 1 1 x dx = lim c →∞ ln x c 1 = lim c →∞ (ln c - 0) = Thus R 1 1 x dx diverges . There is an infinite amount of area under the graph of f ( x ) = 1 x on the interval [1 , ).