Describe the mistakes. Explain your thinking fully and show your work to provide corrections to the mistakes.

c$.$ Calculate ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(2x+1{)}^{2}dx$

Solution:

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(2x+1{)}^{2}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(2x+1)(2x+1)dx$

$=(x^2+x)(x^2+x)+C$

$=(x^2+x{)}^2+C$

Describe the mistakes. Explain your thinking fully and show your work to provide corrections to the mistakes.

d$.$ Find the area enclosed by the line $y=x$ and the curve $y=x^2.$

Solution: Setting the functions equal, we have

$x^2=xso{x}^2-x=0=>x(x-1)=0$

Which happens when $x=0$ and $x=1.$ Then the area between the curves is

${\displaystyle \underset{0}{\overset{1}{}}}(x^2-x)dx=\frac{1}{3}{x}^3-\frac{1}{2}{x}^2{|}_0^1|=\frac{1}{3}(1{)}^3-\frac{1}{2}(1{)}^2=-\frac{1}{6}$

Describe the mistakes. Explain your thinking fully and show your work to provide corrections to the mistakes.

$2.$ In probability and statistics, a probability density function $($or pdif$)$ of a random variable $x$ on the interval $a,b$ is a continuous function $f(x)$ where

$f(x)0$ for all values of $axb$; and

${\displaystyle \underset{a}{\overset{b}{}}}f(x)dx=1$

a$.$ Find a value for $c$ such that the functic $f(x)=c\sqrt[\phantom{2}]{x}$ is a pdf on the interval $(0,4.,$ Show your work completely to justify your response.

b$.$ Consider the function $g(x)=\frac{1}{2}sin(x)$ on $,$ Find the smallest value of be that makes $g(x)$ a pdf on the intervel $,$ b$),$ Show your werk completely to Justify your response.