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Ahat [919]
3 years ago
12

Consider the following theorem. Theorem If f is integrable on [a, b], then b a f(x) dx = lim n→[infinity] n i = 1 f(xi)Δx where

Δx = b − a n and xi = a + iΔx. Use the given theorem to evaluate the definite integral. 9 (x2 − 4x + 6) dx 1
Mathematics
1 answer:
mel-nik [20]3 years ago
3 0

Split up the interval [1, 9] into <em>n</em> subintervals of equal length (9 - 1)/<em>n</em> = 8/<em>n</em> :

[1, 1 + 8/<em>n</em>], [1 + 8/<em>n</em>, 1 + 16/<em>n</em>], [1 + 16/<em>n</em>, 1 + 24/<em>n</em>], …, [1 + 8 (<em>n</em> - 1)/<em>n</em>, 9]

It should be clear that the left endpoint of each subinterval make up an arithmetic sequence, so that the <em>i</em>-th subinterval has left endpoint

1 + 8/<em>n</em> (<em>i</em> - 1)

Then we approximate the definite integral by the sum of the areas of <em>n</em> rectangles with length 8/<em>n</em> and height f(x_i) :

\displaystyle \int_1^9 (x^2-4x+6) \,\mathrm dx \approx \sum_{i=1}^n \frac8n\left(\left(1+\frac8n(i-1)\right)^2-4\left(1+\frac8n(i-1)\right)+6\right)

Take the limit as <em>n</em> approaches infinity and the approximation becomes exact. So we have

\displaystyle \int_1^9 (x^2-4x+6) \,\mathrm dx = \lim_{n\to\infty} \sum_{i=1}^n \frac8n\left(\left(1+\frac8n(i-1)\right)^2-4\left(1+\frac8n(i-1)\right)+6\right) \\\\ = \lim_{n\to\infty} \frac8n \sum_{i=1}^n \left(1+\frac{16}n(i-1)+\frac{64}{n^2}(i-1)^2-4-\frac{32}n(i-1)+6\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \sum_{i=1}^n \left(64(i-1)^2-16n(i-1)+3n^2\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \sum_{i=0}^{n-1} \left(64i^2-16ni+3n^2\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(64\sum_{i=0}^{n-1}i^2 - 16n\sum_{i=0}^{n-1}i + 3n^2\sum{i=0}^{n-1}1\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(\frac{64(2n-1)n(n-1)}{6} - \frac{16n^2(n-1)}{2} + 3n^3\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(\frac{49n^3}3-24n^2+\frac{32n}3\right) \\\\= \lim_{n\to\infty} \frac{8\left(49n^2-72n+32\right)}{3n^2} = \boxed{\frac{392}3}

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Answer:

$ \frac{\sqrt{3} - 1}{2\sqrt{2}} $

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Step-by-step explanation:

Given $ \frac{11 \pi}{12} = \frac{3 \pi}{4} + \frac{\pi}{6} $

(A) $ sin(\frac{11\pi}{12}) = sin (\frac{3 \pi}{4}  + \frac{\pi}{6}) $

We know that Sin(A + B) = SinA cosB + cosAsinB

Substituting in the above formula we get:

$ sin (\frac{3\pi}{4} + \frac{\pi}{6}) = \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} + \frac{-1}{\sqrt{2}}. \frac{1}{2} $

$ \implies \frac{1}{\sqrt{2}} (\frac{\sqrt{3} - 1}{2}) = \frac{\sqrt{3} - 1}{2\sqrt{2}}

(B) Cos(A + B) = CosAcosB - SinASinB

$ cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}}) $

$ \implies \frac{-1}{\sqrt{2}}. \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2} $

$ \implies cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}) $

$ = \frac{-(\sqrt{3} + 1)}{2\sqrt{2}}

(C) Tan(A + B) = $ \frac{Sin(A +B)}{Cos(A + B)} $

From the above obtained values this can be calculated and the value is $ - \frac{\sqrt{3} - 1}{\sqrt{3} + 1} $.

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