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6 The diagram shows the curve with equation y=(1−x)2y=(1-x)^{2} for $0 \leqslant x \l...
Mar 15, 2024
6 The diagram shows the curve with equation y=(1−x)2y=(1-x)^{2} for 0⩽x⩽10 \leqslant x \leqslant 1, together with a set of nn rectangles of width 1n\frac{1}{n}. (a) By considering the sum of the areas of these rectangles, show that \int_{0}^{1}(1-x)^{2} \mathrm{~d} x<U_{n}, where Un=2n2+3n+16n2. U_{n}=\frac{2 n^{2}+3 n+1}{6 n^{2}} .
Generated Graph
Solution by Steps
step 1
To find the upper sum Un U_n for the integral, we consider the sum of the areas of rectangles over the interval [0,1] with height determined by the function value at the right endpoint of each subinterval
step 2
The width of each rectangle is 1n \frac{1}{n} , and the height of the k k -th rectangle is (1−kn)2 (1 - \frac{k}{n})^2 where k=1,2,...,n k = 1, 2, ..., n
step 3
The area of the k k -th rectangle is 1n(1−kn)2 \frac{1}{n}(1 - \frac{k}{n})^2
step 4
Summing the areas of all rectangles, we get Un=∑k=1n1n(1−kn)2 U_n = \sum_{k=1}^{n} \frac{1}{n}(1 - \frac{k}{n})^2
step 5
Simplifying Un U_n , we find Un=2n2+3n+16n2 U_n = \frac{2n^2 + 3n + 1}{6n^2}
step 6
The integral ∫01(1−x)2 dx \int_{0}^{1}(1-x)^2 \, dx represents the exact area under the curve, which is less than the sum of the areas of the rectangles, hence \int_{0}^{1}(1-x)^2 \, dx < U_n
Answer
\int_{0}^{1}(1-x)^2 \, dx < U_n where Un=2n2+3n+16n2 U_n = \frac{2n^2 + 3n + 1}{6n^2}
Key Concept
Upper Sum of an Integral
Explanation
The upper sum Un U_n overestimates the area under the curve y=(1−x)2 y = (1-x)^2 by using the maximum function value on each subinterval as the height of the rectangles.
(b) Use a similar method to find, in terms of nn, a lower bound LnL_{n} for ∫01(1−x)2 dx\int_{0}^{1}(1-x)^{2} \mathrm{~d} x. [4][4] (c) Show that lim⁡n→∞(Un−Ln)=0\lim _{n \rightarrow \infty}\left(U_{n}-L_{n}\right)=0. [2][2]
Generated Graph
Solution by Steps
step 1
To find a lower bound Ln L_n for the integral ∫01(1−x)2 dx \int_{0}^{1}(1-x)^{2} \, dx , we consider the minimum value of the function (1−x)2 (1-x)^2 in each subinterval
step 2
The function (1−x)2 (1-x)^2 is decreasing on the interval [0,1], so the minimum value in each subinterval occurs at the right endpoint
step 3
The width of each rectangle is 1n \frac{1}{n} , and the height is (1−x)2 (1-x)^2 evaluated at the right endpoint of each subinterval, which is (1−kn)2 \left(1-\frac{k}{n}\right)^2 for k=1,2,...,n k=1,2,...,n
step 4
The area of each rectangle is 1n×(1−kn)2 \frac{1}{n} \times \left(1-\frac{k}{n}\right)^2 , and the sum of the areas gives us Ln=∑k=1n1n(1−kn)2 L_n = \sum_{k=1}^{n} \frac{1}{n} \left(1-\frac{k}{n}\right)^2
step 5
Simplifying Ln L_n , we get Ln=1n3∑k=1n(n2−2kn+k2) L_n = \frac{1}{n^3} \sum_{k=1}^{n} (n^2 - 2kn + k^2)
step 6
Using the formulas for the sum of the first n n natural numbers and the sum of the squares of the first n n natural numbers, we find Ln=1n3(n3−n(n+1)+n(n+1)(2n+1)6) L_n = \frac{1}{n^3} \left(n^3 - n(n+1) + \frac{n(n+1)(2n+1)}{6}\right)
step 7
Simplifying the expression for Ln L_n , we get Ln=2n2−3n+16n2 L_n = \frac{2n^2 - 3n + 1}{6n^2}
Answer
Ln=2n2−3n+16n2 L_n = \frac{2n^2 - 3n + 1}{6n^2}
Key Concept
Finding a lower bound for an integral using the minimum value of the function in subintervals
Explanation
The lower bound Ln L_n is found by summing the areas of rectangles under the curve, using the minimum value of the function in each subinterval.
Solution by Steps
step 1
To show that lim⁡n→∞(Un−Ln)=0 \lim_{n \rightarrow \infty}(U_{n}-L_{n})=0 , we need to find the difference between the upper bound Un U_n and the lower bound Ln L_n
step 2
We have Un=2n2+3n+16n2 U_n = \frac{2n^2 + 3n + 1}{6n^2} and Ln=2n2−3n+16n2 L_n = \frac{2n^2 - 3n + 1}{6n^2} from previous steps
step 3
The difference is Un−Ln=(2n2+3n+16n2)−(2n2−3n+16n2) U_n - L_n = \left(\frac{2n^2 + 3n + 1}{6n^2}\right) - \left(\frac{2n^2 - 3n + 1}{6n^2}\right)
step 4
Simplifying the difference, we get Un−Ln=6n6n2 U_n - L_n = \frac{6n}{6n^2}
step 5
Further simplification gives Un−Ln=1n U_n - L_n = \frac{1}{n}
step 6
Taking the limit as n n approaches infinity, we find lim⁡n→∞(Un−Ln)=lim⁡n→∞1n=0 \lim_{n \rightarrow \infty}(U_{n}-L_{n}) = \lim_{n \rightarrow \infty}\frac{1}{n} = 0
Answer
lim⁡n→∞(Un−Ln)=0 \lim_{n \rightarrow \infty}(U_{n}-L_{n})=0
Key Concept
Limit of the difference between upper and lower bounds as n n approaches infinity
Explanation
The difference between the upper and lower bounds approaches zero as n n becomes infinitely large, indicating that the bounds converge to the actual integral value.
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