Use integration by substitution to solve the integral below.

\(\int 4e^{-7x}dx \\\) \(\\u = -7x; du = -7\\\) \(-\frac{1}{7}du = dx\\\) \(-\frac{4}{7}\int e^u du = -\frac{4}{7} e^u + C = -\frac{4}{7} e^{-7x}+ C\)

  1. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of $ =-3150t^{-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.

If the rate of change given DN/Dt is provided then it represents the derivative of the function that is asked for in the question and the best way to determine the function is to take the integral of the rate.

\(\frac{dN}{dt} = -\frac{3150}{t^4} - 220\) \(\int -\frac{3150}{t^4} - 220 dt = \frac{3150t^{-3}}{3}-220t+C\) \(\frac{3150t^{-3}}{3}-220t+C = 6530 |t=1 = 3150*1^{-3}*3^{-1}-220*-1+C = 6530\) \(-\frac{3150}{1*3}+220-6530=-C; C=5700\) \(N(t) = \frac{3150t^{-3}}{3}-220t+5700\)

# Calculating constant
3150*(1/3)*(1^3)-220*(1)-6530
## [1] -5700
  1. Find the total area of the red rectangles in the figure below, where the equation of the line is f(x) = 2x - 9.

The banded area on the x plane is 9/2 through 17/2 \(\int_{5.5}^{8.5} 2x-9 dx\) \(\left. x^2-9x\right|_{4.5}^{8.5}\)

func_x <- function(x) {
    2*x-9
    }
integrate(func_x,lower=9/2,upper=17/2)
## 16 with absolute error < 1.8e-13
(8.5)**2-9*8.5 - ((4.5)**2-9*(4.5) )
## [1] 16

Area = 16 and both methods confirm the integral of the function 2x-9 between 4.5 and 8.5

  1. Find the area of the region bounded by the graphs of the given equations. \(y=x^2-2x-2, y=x+2\) In order to properly determine the area of the two below functions we need to set them equal to one another to determine the points of intersection

\(x^2-2x-2=x+2;\) \(x^2-3x-4 = (x-4)(x+1)\)

The two functions intersect at x= 4 and -1 and we can easily determine that the upper bound function is x+2 because at zero it returns a higher value

\(x+2 - (x^2-2x-2) = x+2-x^2+2x+2 = \int_{-1}^{4}3x-x^2+4 dx\) \(\left. \frac{3x^2}{2}-\frac{x^3}{3}+4x\right|_{-1}^{4}\)

-4^3/3+(3*4^2)/2+4*4 - (-(-1)^3/3+(3*(-1)^2)/2+(-1)*4)
## [1] 20.83333

Enter your answer below.

Estimated area between these two curves is 20.833

  1. 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.

Cost(x) = Order Cost + Storage Cost Lot size = x Average irons in storage = x/2 Number of orders = n xn = 110 x = 110/n

\(c(x) = 3.75 \frac{110/n}{2}+8.25*n = \frac{55*15}{4n}+8.25n = 55*15*\frac{1}{4}n^{-1}+8.25n\) \(c\prime(x) = -55*15*\frac{1}{4}n^{-2}+8.25\) Set equal to zero to find the minimum \(0 = -55*15*\frac{1}{4}n^{-2}+8.25; 8.25n^2 = 55*15*\frac{1}{4} = \sqrt{\frac{55*15}{4*8.25}}\) \(n=5\)

sqrt((55*15)/(4*8.25))
## [1] 5
3.75*22/2+8.25*5
## [1] 82.5

Therefore the number of orders per year is minimized with 5 orders with a lot size of 22 equating to a cost of $82.50

  1. Use integration by parts to solve the integral below.

\(\int ln(9x) x^6 dx\)

Substitution by parts integration formula: \(\int u dv = uv - \int vdu\)

\(u = ln(x)\) \(du = \frac{1}{x}\) \(dv = x^6; v = \int x^6 dx =\frac{x^7}{7}\) \(v = \frac{x^7}{7}\)

\(ln(9x)* \frac{x^7}{7} - \int \frac{x^7}{7}*\frac{1}{x} dx\) \(ln(9x)* \frac{x^7}{7} - \int \frac{x^6}{7} dx\) \(ln(9x)* \frac{x^7}{7} - \frac{x^7}{49} + C\)

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

\(f(x)= \frac{1}{6x}\)

The first assumption for a pdf is that it cannot be a negative value and since the range is above zero and all other values would be zero it is sufficient to satisfy this criteria.

The second criteria that is needed for a valid pdf is that the sum of the probabilities is equal to 1. Therefore taking the integral between these ranges will be needed to assess the area.

\(\int_{-\infty}^{+\infty} f(x) dx = \int_{1}^{e^6} \frac{1}{6x} dx = \frac{1}{6}\int_{1}^{e^6} \frac{1}{x} dx\)

\(\left.-\frac{1}{6}ln(x) \right|_{1}^{e^6}\)

\(\frac{1}{6}ln(e^6) - \frac{1}{6}ln(1)\)

(1/6)*log(exp(6))-(1/6)*log(1)
## [1] 1

Since the integral represents the area under this curve and it evaluates to 1 over this range of values this does represent a probability density function after proving this criteria