How do you find the average value of the function for $f(x)=\sqrt{x}+\frac{1}{\sqrt{x}},1\le x\le 9$?

civilnogwu
2022-07-07
Answered

How do you find the average value of the function for $f(x)=\sqrt{x}+\frac{1}{\sqrt{x}},1\le x\le 9$?

You can still ask an expert for help

Zackery Harvey

Answered 2022-07-08
Author has **21** answers

Use the formula $\frac{1}{b-a}{\int}_{a}^{b}f(x)dx$ and the Fundamental Theorem of Calculus (FTC) to get an average value equal to $\frac{8}{3}$.

Explanation:

The average value is $\frac{1}{b-a}{\int}_{a}^{b}f(x)dx=\frac{1}{9-1}{\int}_{1}^{9}({x}^{\frac{1}{2}+{x}^{-\frac{1}{2}}})dx$.

The FTC then leads us to say this equals

$\frac{1}{8}(\frac{2}{3}{x}^{\frac{3}{2}}+2{x}^{\frac{1}{2}}){|}_{1}^{9}=\frac{1}{8}((\frac{2}{3}\ast 27+2\ast 3)-(\frac{2}{3}\ast 1+2\ast 1))$

This simplifies to

$\frac{1}{8}(24-\frac{8}{3})=\frac{1}{8}(\frac{64}{3})=\frac{8}{3}$

Explanation:

The average value is $\frac{1}{b-a}{\int}_{a}^{b}f(x)dx=\frac{1}{9-1}{\int}_{1}^{9}({x}^{\frac{1}{2}+{x}^{-\frac{1}{2}}})dx$.

The FTC then leads us to say this equals

$\frac{1}{8}(\frac{2}{3}{x}^{\frac{3}{2}}+2{x}^{\frac{1}{2}}){|}_{1}^{9}=\frac{1}{8}((\frac{2}{3}\ast 27+2\ast 3)-(\frac{2}{3}\ast 1+2\ast 1))$

This simplifies to

$\frac{1}{8}(24-\frac{8}{3})=\frac{1}{8}(\frac{64}{3})=\frac{8}{3}$

asked 2022-07-14

Integral of $\frac{\sqrt{x}}{{x}^{2}+x}$

asked 2022-07-14

How do can you derive the equation for a circle's circumference using integration?

asked 2022-04-18

What is a solution to the differential equation $\frac{dy}{dx}=x{e}^{y}$ ?

asked 2022-04-11

How do you solve separable first-order differential equations?

asked 2022-05-20

Integral of a trig function divided by the square root of a polynomial: ${\int}_{a}^{b}\frac{\mathrm{sin}x}{\sqrt{(x-a)(b-x)}}dx$

asked 2021-12-27

Find the indefinite integral.

$\int (2x+1){e}^{{x}^{2}+x}dx$

asked 2022-01-01

Trigonometric integrals Evaluate the following integrals.

$\int {\mathrm{tan}}^{2}xdx$