530
Chapter 5 | Integration
→∞ ∑ i =1 n
f ⎛
⎞ ⎠ Δ x exists and is
Previously, we discussed the fact that if f ( x ) is continuous on ⎡ ⎣ a , b ⎤
⎦ , then the limit lim n
⎝ x i *
unique. This leads to the following theorem, which we state without proof.
Theorem 5.1: Continuous Functions Are Integrable If f ( x ) is continuous on ⎡ ⎣ a , b ⎤ ⎦ , then f is integrable on ⎡ ⎣ a , b ⎤ ⎦ .
Functions that are not continuous on ⎡
⎤ ⎦ may still be integrable, depending on the nature of the discontinuities. For
⎣ a , b
example, functions with a finite number of jump discontinuities on a closed interval are integrable. It is also worth noting here that we have retained the use of a regular partition in the Riemann sums. This restriction is not strictly necessary. Any partition can be used to form a Riemann sum. However, if a nonregular partition is used to define the definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must take the limit as the width of the largest subinterval goes to zero. This introduces a little more complex notation in our limits and makes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions for the Riemann sums. Example 5.7 Evaluating an Integral Using the Definition
Use the definition of the definite integral to evaluate ∫ 0 2
x 2 dx . Use a right-endpoint approximation to generate
the Riemann sum.
Solution We first want to set up a Riemann sum. Based on the limits of integration, we have a =0 and b =2. For i =0, 1, 2,…, n , let P ={ x i } be a regular partition of [0, 2]. Then Δ x = b − a n = 2 n . Since we are using a right-endpoint approximation to generate Riemann sums, for each i , we need to calculate the function value at the right endpoint of the interval [ x i −1 , x i ]. The right endpoint of the interval is x i , and since P is a regular partition, x i = x 0 + i Δ x =0+ i ⎡ ⎣ 2 n ⎤ ⎦ = 2 i n . Thus, the function value at the right endpoint of the interval is f ( x i ) = x i 2 = ⎛ ⎝ 2 i n ⎞ ⎠ 2 = 4 i 2 .
n 2
Then the Riemann sum takes the form ∑ i =1 n Using the summation formula for ∑ i =1 n
n ⎛
⎞ ⎠ 2
n = ∑ i =1 n
∑ i =1 n
8 i 2 n 3
⎝ 4 i 2 n 2
f ( x i )Δ x = ∑ i =1
i 2 .
= 8
n 3
i 2 , we have
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