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2. (a) For integer n≥1n \geq 1, use upper and lower Riemann sums with nn equal ...
May 5, 2024
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Solution by Steps
step 1
To find the upper and lower Riemann sums for the integral ∫01(1−x2)dx\int_{0}^{1}(1-x^{2})dx, we divide the interval [0,1] into nn equal subintervals of width Δx=1n\Delta x = \frac{1}{n}
step 2
The upper sum UnU_n is found by evaluating (1−x2)(1-x^2) at the right endpoint of each subinterval, which gives us Un=∑k=1n(1−(kn)2)1nU_n = \sum_{k=1}^{n} \left(1-\left(\frac{k}{n}\right)^2\right)\frac{1}{n}
step 3
Using the formula for the sum of squares, we have Un=∑k=1n1n−∑k=1nk2n3=1−(n)(n+1)(2n+1)6n3U_n = \sum_{k=1}^{n} \frac{1}{n} - \sum_{k=1}^{n} \frac{k^2}{n^3} = 1 - \frac{(n)(n+1)(2n+1)}{6n^3}
step 4
Simplifying UnU_n, we get Un=1−(n+1)(2n+1)6n2U_n = 1 - \frac{(n+1)(2n+1)}{6n^2}
step 5
The lower sum LnL_n is found by evaluating (1−x2)(1-x^2) at the left endpoint of each subinterval, which gives us Ln=∑k=0n−1(1−(kn)2)1nL_n = \sum_{k=0}^{n-1} \left(1-\left(\frac{k}{n}\right)^2\right)\frac{1}{n}
step 6
Using the formula for the sum of squares with a shift in index, we have Ln=∑k=0n−11n−∑k=0n−1k2n3=1−(n−1)(n)(2n−1)6n3L_n = \sum_{k=0}^{n-1} \frac{1}{n} - \sum_{k=0}^{n-1} \frac{k^2}{n^3} = 1 - \frac{(n-1)(n)(2n-1)}{6n^3}
step 7
Simplifying LnL_n, we get Ln=1−(n)(2n−1)6n2L_n = 1 - \frac{(n)(2n-1)}{6n^2}
step 8
To evaluate the integral using the definition of the Riemann integral, we take the limit of the Riemann sums as nn approaches infinity
step 9
The limit of the upper sum as nn approaches infinity is lim⁡n→∞Un=lim⁡n→∞(1−(n+1)(2n+1)6n2)=13\lim_{n \to \infty} U_n = \lim_{n \to \infty} \left(1 - \frac{(n+1)(2n+1)}{6n^2}\right) = \frac{1}{3}
step 10
The limit of the lower sum as nn approaches infinity is lim⁡n→∞Ln=lim⁡n→∞(1−(n)(2n−1)6n2)=13\lim_{n \to \infty} L_n = \lim_{n \to \infty} \left(1 - \frac{(n)(2n-1)}{6n^2}\right) = \frac{1}{3}
step 11
Since the upper and lower sums converge to the same value, the value of the integral is 13\frac{1}{3}
[1] Answer
The upper and lower bounds for the integral are both 13\frac{1}{3}, hence the value of the integral is 13\frac{1}{3}.
Key Concept
Riemann sums and definite integrals
Explanation
The upper and lower Riemann sums provide bounds for the value of a definite integral. As the number of subdivisions increases, both sums converge to the actual value of the integral if the function is integrable on the interval.
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