Not entirely sure, but it seem the answer could be B.) It will not be spread out vertically across the entire coordinate plane because in step 5, Nancy selected an incorrect scale on the y-axis.

3 0

3 years ago

Read 2 more answers

Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your

Darina [25.2K]

Answer:

Given definite integral as a limit of Riemann sums is:

$\lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

Step-by-step explanation:

Given definite integral is:

$\int\limits^7_4 {\frac{x}{2}+x^{3}} \, dx \\f(x)=\frac{x}{2}+x^{3}---(1)\\\Delta x=\frac{b-a}{n}\\\\\Delta x=\frac{7-4}{n}=\frac{3}{n}\\\\x_{i}=a+\Delta xi\\a= Lower Limit=4\\\implies x_{i}=4+\frac{3}{n}i---(2)\\\\then\\f(x_{i})=\frac{x_{i}}{2}+x_{i}^{3}$

Substituting (2) in above

$f(x_{i})=\frac{1}{2}(4+\frac{3}{n}i)+(4+\frac{3}{n}i)^{3}\\\\f(x_{i})=(2+\frac{3}{2n}i)+(64+\frac{27}{n^{3}}i^{3}+3(16)\frac{3}{n}i+3(4)\frac{9}{n^{2}}i^{2})\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{3}{2n}i+\frac{144}{n}i+66\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{291}{2n}i+66\\\\f(x_{i})=3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

Riemann sum is:

$= \lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

4 0

4 years ago

What is the value of x????

Arte-miy333 [17]

Answer:

Step-by-step explanation:

${18}^{2} = {x}^{2} + {x}^{2} \\ \frac{18 \times 18}{2} = {x}^{2} \\ x = \sqrt{9 \times 18} \\ x = \sqrt{9 \times 9 \times 2} \\ x = 9 \sqrt{2}$

4 0

3 years ago