Mathematics, 03.08.2021 15:50 nsaleh567

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

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Answer from: thetudent41

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

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

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

1 + 8/n (i - 1)

Then we approximate the definite integral by the sum of the areas of n rectangles with length 8/n 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 n approaches infinity and the approximation becomes exact. So we have

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

Answer from: Quest

x= 1

step-by-step explanation: convert 5\frac{11}{12}5

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{5\times 12+11}{12}3.25+0.5x+1

5×12+11

2 simplify 5\times 125×12 to 6060

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{60+11}{12}3.25+0.5x+1

60+11

3 simplify 60+1160+11 to 7171

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{71}{12}3.25+0.5x+1

4 simplify \frac{5}{6}x

x to \frac{5x}{6}

3.25+0.5x+1\frac{1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+1

5 convert 1\frac{1}{3}1

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+\frac{1\times 3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

1×3+1

6 simplify 1\times 31×3 to 33

3.25+0.5x+\frac{3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

3+1

7 simplify 3+13+1 to 44

3.25+0.5x+\frac{4}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

8 simplify 3.25+0.5x+\frac{4}{3}+\frac{5x}{6}3.25+0.5x+

to \frac{13.75}{3}+\frac{4x}{3}

13.75

\frac{13.75}{3}+\frac{4x}{3}=\frac{71}{12}

13.75

9 simplify \frac{13.75}{3}

13.75

to 4.5833334.583333

4.583333+\frac{4x}{3}=\frac{71}{12}4.583333+

10 subtract 4.5833334.583333 from both sides

\frac{4x}{3}=\frac{71}{12}-4.583333

−4.583333

11 simplify \frac{71}{12}-4.583333

−4.583333 to \frac{4}{3}

\frac{4x}{3}=\frac{4}{3}

12 multiply both sides by 33

4x=\frac{4}{3}\times 34x=

×3

13 simplify \frac{4}{3}\times 3

×3 to \frac{12}{3}

4x=\frac{12}{3}4x=

14 simplify \frac{12}{3}

to 44

4x=44x=4

15 divide both sides by 44

x=1

Answer from: Quest

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