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| | |Finally, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit. | | |Finally, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit. |
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| − | |Thus, the area of <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> is equal to <math style="vertical-align: -21px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)</math>. | + | |Thus, <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> is equal to <math style="vertical-align: -21px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)</math>. |
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Revision as of 08:33, 1 February 2016
Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=1-x^2}
.
- a) Compute the left-hand Riemann sum approximation of
with
boxes.
- b) Compute the right-hand Riemann sum approximation of
with
boxes.
- c) Express
as a limit of right-hand Riemann sums (as in the definition of the definite integral). Do not evaluate the limit.
| Foundations:
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| Link to Riemann sums page
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Solution:
(a)
| Step 1:
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Since our interval is and we are using 3 rectangles, each rectangle has width 1. So, the left-hand Riemann sum is
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.
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| Step 2:
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| Thus, the left-hand Riemann sum is
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.
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(b)
| Step 1:
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Since our interval is and we are using 3 rectangles, each rectangle has width 1. So, the right-hand Riemann sum is
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.
|
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|
| Step 2:
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| Thus, the right-hand Riemann sum is
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.
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(c)
| Step 1:
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Let be the number of rectangles used in the right-hand Riemann sum for .
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The width of each rectangle is .
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| Step 2:
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| So, the right-hand Riemann sum is
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| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \Delta x{\bigg (}f{\bigg (}1\cdot {\frac {3}{n}}{\bigg )}+f{\bigg (}2\cdot {\frac {3}{n}}{\bigg )}+f{\bigg (}3\cdot {\frac {3}{n}}{\bigg )}+\ldots +f(3){\bigg )}}
.
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Finally, we let go to infinity to get a limit.
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Thus, is equal to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)}
.
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| Final Answer:
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| (a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -2}
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| (b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -11}
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| (c) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)}
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