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goldfiish [28.3K]
3 years ago
15

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Mathematics
2 answers:
Vinvika [58]3 years ago
6 0
The answer is blue 17 3
ss7ja [257]3 years ago
3 0

Answer:

Step-by-step explanation:

20

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Harlamova29_29 [7]

The summand (R?) is missing, but we can always come up with another one.

Divide the interval [0, 1] into n subintervals of equal length \dfrac{1-0}n=\dfrac1n:

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Let's consider a left-endpoint sum, so that we take values of f(\ell_i)={\ell_i}^3 where \ell_i is given by the sequence

\ell_i=\dfrac{i-1}n

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\displaystyle\int_0^1x^3\,\mathrm dx=\lim_{n\to\infty}\sum_{i=1}^n\left(\frac{i-1}n\right)^3\frac{1-0}n

=\displaystyle\lim_{n\to\infty}\frac1{n^4}\sum_{i=1}^n(i-1)^3

=\displaystyle\lim_{n\to\infty}\frac1{n^4}\sum_{i=0}^{n-1}i^3

=\displaystyle\lim_{n\to\infty}\frac{n^2(n-1)^2}{4n^4}=\boxed{\frac14}

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4 years ago
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Which statement show equivalent expressions? PLEASE HELP ME ASAP
Zigmanuir [339]

Answer:

c.

Step-by-step explanation:

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3 years ago
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nadezda [96]

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