Assignment

Question 1:

Use integration by substitution to solve the integral below:

\[\int { 4{ e }^{ -7x } } dx\] \[u=\frac { du }{ dx } =-7\] \[du=-7dx\] \[dx=\frac { du }{ -7 } =-\frac { 1 }{ 7 } du\] \[-\frac { 4 }{ 7 } \int { { e }^{ u } } du\] \[\frac { 4 }{ 7 } { { e }^{ u } }=\frac { 4 }{ 7 } { e }^{ -7x }+c\]

Question 2

Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \(\frac { dN }{ dt } =\frac { 3150 }{ { t }^{ 4 } } -220\) bacteria per cubic centimeter per day, where t is the number of days since treatment began. Find a function N( t ) to estimate the level of contamination if the level after 1 day was 6530 bacteria per cubic centimeter.

\[\frac { dN }{ dt } =\frac { 3150 }{ { t }^{ 4 } } -220\] \[dN=\frac { 3150 }{ { t }^{ 4 } } -220dt\] \[dN=\int { \frac { 3150 }{ { t }^{ 4 } } } dt-\int { 220dt } \] \[dN=\int { -3150{ t }^{ -4 } } dt-\int { 220dt } \] \[N\left( t \right) =\frac { 3150 }{ 3{ t }^{ 4 } } -220t\quad +C\] \[time(t)=1\] \[N(1)[6530\_ bacteria\_ per\_ { c }^{ 3 }]=6350\] \[N\left( t \right) =\frac { -3150 }{ -3 } -220+C=6350\] \[6530=1050{ (1) }^{ -3 }-220(1)+C\] \[C=5700\]

C<- 6530+3150*(1^3)/-3+220*(1)
C
## [1] 5700

\[N(t)=1050{ t }^{ -3 }-220t+5700\]

Question 3

Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x)=2x-9\).

\[A=\int _{ 4.5 }^{ 8.5 }{ (2x-9)dx } \]
\[A=({ 8.5 }^{ 2 }-9(8.5)-({ 4.5 }^{ 2 }-9(4.5)\] Lets let R do the work for us this time.

Q3<- 8.5^2-(9*8.5)-(4.5^2-(9*4.5))
Q3
## [1] 16

Question 4

Find the area of the region bounded by the graphs of the given equations:

\[y={ x }^{ 2 }-2x-2,y=x+2\]

fun1 <- function(x)
{
  x^2-2*x-2
}
fun2 <- function(x)
{
  x+2
}

curve(fun1,-7,7)
curve(fun2, -7,7, add = TRUE)

\[{ x }^{ 2 }-2x-2=x+2\] \[{ x }^{ 2 }-3x-4=0\] \[(x-4)(x+1)=0\] X values become 4,-1 y values become 6,1 making the points (4,6),(1-,1) \[\int _{ -1 }^{ 4 }{ x } +2dx-\int _{ -1 }^{ 4 }{ { x }^{ 2 } } -2x-2dx\] \[\int _{ -1 }^{ 4 }{ x } -\left\{ \frac { 1 }{ 3 } { x }^{ 3 }-\frac { 3 }{ 2 } { x }^{ 2 }-4x \right\} \] \[A=20.83\]

Question 5

A beauty supply store expects to sell 110 flat irons during the next year. It costs $3.75 to store one flat iron for one year. There is a fixed cost of $8.25 for each order. Find the lot size and the number of orders per year that will minimize inventory costs.

Definations:
x=number of orders y=Size of lot S=Storage of 1 iron C=Cost of 1 iron

Knowns: Total number of irons ordered for Year = x*y minimum of 1 order per year so x>=1 Expect to sell more than 110 so x is also x>=110 S=3.75 C=8.25

Formula: \[{ C }_{ t }=8.25n+\frac { 3.75 }{ 2n } \] \[\acute { { C }_{ t } } =8.25-\frac { 206.25 }{ { n }^{ 2 } } \] \[0=8.25-\frac { 206.25 }{ { n }^{ 2 } } \] \[n=5\] \[110/5=22\]

The store will need to order 22 units 5 times through out the year to achieve 110 sales at the minimal inventory cost.

QUestion 6

Use integration by parts to solve the integral below.

\[\int { ln } \left( 9x \right) *{ x }^{ 6 }dx\]

The formula in Question is: \[\int { f\left( x \right) g^{ , }\left( x \right) } =f\left( x \right) g\left( x \right) -\int { f^{ ' }\left( x \right) } g\left( x \right)\] First we take apart the first section and simplify \[ln(9x)=\frac { 9 }{ 9x } dx=\frac { 1 }{ x } dx\] Second, we take apart the second section and simplify \[{ x }^{ 6 }dx<\equiv >\frac { dv }{ dx } ={ x }^{ 6 }<\equiv >dv={ x }^{ 6 }dx\] \[v=\frac { 1 }{ 7 } { x }^{ 7 }\] Plugging those into the formula Creates: \[ln(9x)\frac { { x }^{ 7 } }{ 7 } -\frac { { x }^{ 7 } }{ 49 } +C\]

Question 7

Determine whether \(f\left( x \right)\) is a probability density function on the interval $$. If not, determine the value of the definite integral.

\[f\left( x \right) =\frac { 1 }{ 6x } \] \[\int _{ 1 }^{ { e }^{ 6 } }{ \frac { 1 }{ 6x } } dx\] \[\frac { 1 }{ 6 } \int _{ 1 }^{ { e }^{ 6 } }{ \frac { 1 }{ x } } dx\] \[\frac { 1 }{ 6 } ln(x)|\begin{matrix} { e }^{ 6 } \\ 1 \end{matrix}\] \[\frac { 1 }{ 6 } \left[ ln({ e }^{ 6 })-ln(1) \right] \] \[\frac { 1 }{ 6 } \left[ 6-0 \right] \] \[=1\]