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just olya [345]
3 years ago
12

Please help me!!!! on question 11

Mathematics
1 answer:
marishachu [46]3 years ago
4 0

Answer:

D)

Step-by-step explanation:

y=\frac{-3}{2}x-4

Coefficient of x is -3/2 and so multiply the whole equation by 2

2*y=2*\frac{-3}{2}x-2*4\\\\2y=-3x-8

3x + 2y = -8

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Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your
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The expression of integral as a limit of Riemann sums of given integral \int\limits^5_b {1} \, x/(2+x^{3}) dxis 4 \lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3} from i=1 to i=n.

Given an integral \int\limits^5_b {1} \, x/(2+x^{3}) dx.

We are required to express the integral as a limit of Riemann sums.

An integral basically assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinite data.

A Riemann sum is basically a certain kind of approximation of an integral by a finite sum.

Using Riemann sums, we have :

\int\limits^b_a {f(x)} \, dx=\lim_{n \to \infty}∑f(a+iΔx)Δx ,here Δx=(b-a)/n

\int\limits^5_1 {x/(2+x^{3}) } \, dx=f(x)=x/2+x^{3}

⇒Δx=(5-1)/n=4/n

f(a+iΔx)=f(1+4i/n)

f(1+4i/n)=[n^{2}(n+4i)]/2n^{3}+(n+4i)^{3}

\lim_{n \to \infty}∑f(a+iΔx)Δx=

\lim_{n \to \infty}∑n^{2}(n+4i)/2n^{3}+(n+4i)^{3}4/n

=4\lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3}

Hence the expression of integral as a limit of Riemann sums of given integral \int\limits^5_b {1} \, x/(2+x^{3}) dxis 4 \lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3} from i=1 to i=n.

Learn more about integral at brainly.com/question/27419605

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