# Week7 - 110.109 CALCULUS II Week 7 Lecture Notes March 14...

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Unformatted text preview: 110.109 CALCULUS II Week 7 Lecture Notes: March 14 - March 18 Improper Integrals (cont’d.) Go back to the example of an integral where the integrands is f ( x ) = 1 x 2 , but this time choose to integrate on the interval [ b, 1], where 0 < b ≤ 1. For a choice of b here, we get ∫ 1 b 1 x 2 dx = − 1 x 1 b = − 1 + 1 b > . We know it is greater than 0 since the entire graph of f ( x ) > 0, and also for the last expression 1 b > 1 (why?) Question 1. What if we pushed b all the way back to ? Again, we would produce a problem with the limits of the integral. And again, the problem would involve an interpretation of the definite integral as the area of an unbounded region. However, this time, the region is unbounded in the vertical direction due to the vertical asymptote at x = 0. To deal with this situation, we employ the same trick as before, noting that the definite integral is perfectly well-defined for all positive values of b : define ∫ 1 1 x 2 dx = lim b → + ∫ 1 b 1 x 2 dx = lim b → + ( − 1 x 1 b ) = lim b → + ( − 1 + 1 b ) = lim b → + ( 1 − b b ) . Finishing this calculation means evaluating the limit at the end of the equations above. We will get ∞ , and can interpret this as the area between f ( x ) and the x-axis grows without bound as we push the lower limit back to 0. This is an example of another type of improper integral, where the integrand is not defined at one or both of the limits. Example 2. Do the same thing by replacing the integrand with the new function g ( x ) = 1 3 √ x . Note that the graph look remarkably similar, and there is again a vertical asymptote of g ( x ) at x = 0. However, this time, we get ∫ 1 1 3 √ x dx = lim b → + ∫ 1 b 1 3 √ x dx = lim b → + ( 3 2 x 2 3 1 b ) = lim b → + ( 3 2 − 3 2 3 √ b 2 ) = 3 2 . Here the resulting limit does exist (the function 3 √ x 2 is continuous from the right at x = 0). The interpre- tation is that the area of the unbounded region between the curve and the x-axis is finite and is 3 2 . Example 3. How about h ( x ) = ( x +1)( x- 2) x- 2 on the interval [1 , 3] ? Here both limits of the integral of h ( x ) would be fine. However, there is a point inside the interval (namely x = 2) where the function is not defined. We cannot simply ignore this point. We can, however, adjust the calculation to accommodate it knowing the Sum Law for Integrals (Section 5.2, page 374). We can write ∫ 3 1 ( x + 1)( x − 2) x − 2 dx = ∫ 2 1 ( x + 1)( x − 2) x − 2 dx + ∫ 3 2 ( x + 1)( x − 2) x − 2 dx, knowing that both of the integrals on the right-hand side are improper. Keep in mind that they are improper because one of their limits is outside of the domain of the function, and not because there is an asymptote Date : March 23, 2011....
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Week7 - 110.109 CALCULUS II Week 7 Lecture Notes March 14...

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