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Sergeeva-Olga [200]
3 years ago
12

Suppose that the function g is defined, for all real numbers, as follows. See attached.

Mathematics
1 answer:
tekilochka [14]3 years ago
3 0

(a) We use the given g(x) to find the values

g(-1), Here x = -1 which is x<=-2

so we use g(x) = \frac{1}{4} x - 1

Plug in -1 for x

so g(-1) =  \frac{1}{4}(-1) - 1=  \frac{-1}{4} - 1=  \frac{-5}{4}

g(-1) =  \frac{-5}{4}

(b) g(1), Here x=1  and it comes under the interval x>=1

so we use g(x) = -2

g(1) = -2

(c) g(4), Here x=4  and it comes under the interval x>=1 because 4 is greater than 1. so we use g(x) = -2

g(4) = -2

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4 0
4 years ago
Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your
Veronika [31]

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

#SPJ4

5 0
2 years ago
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