N Selina Assignment 13 - Data 605

Noori Selina

Use integration by substitution to solve the integral below. \[ \int 4e^{-7x} \, dx \]

$u = -7x$, then $du = -7dx$. In terms of $dx$, $dx = -\frac{1}{7} du$.

First we have to rewrite the integral in terms of $u$:

\[ \int 4e^u(-\frac{1}{7}) du \]

Then we will take out constants:

\[ -\frac{4}{7} \int e^u du \]

Integrate:

\[ -\frac{4}{7}(e^u + C) \]

Then we will substitute back in terms of $x$:

\[ -\frac{4}{7}(e^{-7x} + C) \]

For some constant $C$.

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.

\[ \int \left( -\frac{3150}{t^4} - 220 \right) \, dt \]

First we split integral:

\[ \int \left( -\frac{3150}{t^4} \, dt \right) - \int (220 \, dt) \]

Next we have to take out constants for both integrals:

\[ -3150 \int \left( \frac{1}{t^4} \, dt \right) - \int (220 \, dt) \]

Integrate:

\[ -3150 \left( -\frac{1}{3t^3} \right) - 220t \]

Now, We simplify:

\[ \frac{1050}{t^3} - 220t \]

Write as function $N(t)$:

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

For some constant $C$.

Solve for $N(t) = 6530$ where $t = 1$:

\[ \frac{1050}{t^3} - 220t + C = 6530 \] \[ \frac{1050}{1^3} - 220(1) + C = 6530 \] \[ 1050 - 220 + C = 6530 \] \[ 830 + C = 6530 \] \[ C = 5700 \]

Function $N(t) = \frac{1050}{t^3} - 220t + 5700$.

Find the total area of the red rectangles in the figure below, where the equation of the line is f ( x ) = 2x - 9. The rectangles start at $x = 4.5$ and end at $x = 8.5$.

x2−2x−2=x+2 Set equal to 0.

x2−3x−4=0 Factor.

(x−4)(x+1)=0 Solve for x.

x=4,x=−1

Subtract the two equations.

x+2−(x2−2x−2) Simplify.

x+2−x2+2x+2 Combine the like terms.

−x2+3x+4

Then we integrate this equation using intersection points as interval.

\[ \int_{8.5}^{4.5} 2x - 9 \, dx = \left[ x^2 - 9x \right]_{8.5}^{4.5} \]

Area $= (8.5^2 - 9(8.5)) - (4.5^2 - 9(4.5))$

Area $= (72.25 - 76.5) - (20.25 - 40.5)$

Area $= (-4.25) - (-20.25)$

Area = 16

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

y=x2−2x−2,y=x+2

function_4 <- function(x)
{
  -x ^ 2 + 3 * x + 4
}
integrate(function_4, lower = -1, upper = 4)

## 20.83333 with absolute error < 2.3e-13

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.

Number of orders ($x$) * Lot Size ($n$) = 110 C= Costs

\[ C = 8.25x + 3.75 \cdot \left( \frac{110}{x^2} \right) \]

\[ C = 8.25x + 3.75 \cdot \left( \frac{55}{x} \right) \]

\[ C = 8.25x + \frac{206.25}{x} \]

To minimize costs, we need to set the derivative to zero

\[ C' = 8.25 - \frac{206.25}{x^2} \] \[ 0 = 8.25 - \frac{206.25}{x^2} \]

\[ \frac{206.25}{x^2} = 8.25 \] \[ 206.25 = 8.25x^2 \] \[ \frac{206.25}{8.25} = x^2 \] \[ 25 = x^2 \]

The number of orders per year that will minimize inventory costs is 5. The lot size is 22 because $22 \times 5 = 110$.

Use integration by parts to solve the integral below.

\[ \int \ln(9x) \cdot x^6 \, dx \]

$u = \ln(9x)$, and $du = \frac{1}{x} dx$. and $v = \frac{1}{7}x^7$, then $dv = x^6 dx$.

Using integration by parts, we can calculate $u \cdot v - \int v \cdot du$.

\[ \ln(9x) \cdot \frac{1}{7}x^7 - \int \frac{1}{7}x^7 \cdot \frac{1}{x} \, dx \]

Simplifying:

\[ \frac{x^7 \ln(9x)}{7} - \int \frac{x^6}{7} \, dx \]

Integrate:

\[ \frac{x^7 \ln(9x)}{7} - \left( \frac{x^7}{49} + C \right) \]

For some constant $C$.

Determine whether f ( x ) is a probability density function on the interval [1, e ^ 6] . If not, determine the value of the definite integral.

$f( x ) = $

function_7 <- function(x)
{
  1 / (6 * x)
}
integrate(function_7, lower = 1, upper = exp(6))

## 1 with absolute error < 9.3e-05

f(x) is a probability density function on the interval [1, e ^ 6] because the integral of f(x) on the interval = 1.