1. Use integration by substitution to solve the integral below:

\(\int 4e^{-7x} \mathrm{d}x\)

So we need the integral to be in the form

\(\int f(g(x)) g'(x) \mathrm{d}x\)

We can pull the 4 out as a first step

\(4 \int e^{-7x} \mathrm{d}x\)

Then we need to the g’(x):

\(\frac{4}{-7} \int -7e^{-7x} \mathrm{d}x\)

Now our u substitution:

\(\frac{4}{-7} \int e^{u} \mathrm{d}u\)

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

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

  1. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \(\frac{dN}{dt} = - \frac{3150}{t^4} -220\) 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

We can use integration and the power rule to find our equation.

\(f(t) = \int- \frac{3150}{t^4}\mathrm{d}t -\int220\mathrm{d}t\)

\(f(t) = 3150\int- \frac{1}{t^4}\mathrm{d}t -\int220\mathrm{d}t\)

\(f(t) = 3150 \frac{1}{t^3} - 220t + C\)

\(6530 = 3150- 220 + 3600\)

\(f(t) = 3150 \frac{1}{t^3} - 220t + 3600\)

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

I used this page as a reference

first calculate our 4 midpoints on the x axis

x1 = 1/8 * 4.5 + 4.5

x2 = 3/8 * 4.5 + 4.5

x3 = 5/8 * 4.5 + 4.5

x4 = 7/8 * 4.5 + 4.5

find the heights by plugging our midpoints into the equation

h1 = 2*x1 - 9

h2 = 2*x2 - 9

h3 = 2*x3 - 9

h4 = 2*x4 - 9

find our areas by multiplying our width (\(\Delta x\)) by heights and then summing

width = (8.5 - 4.5) * 1/4
a1 = width * h1

a2 = width* h2

a3 = width * h3

a4 = width * h4

total_area = a1 + a2 + a3 + a4

total_area
## [1] 18
  1. Find the area of the region bounded by the graphs of the given equations. \(y = x^2 - 2x - 2\) \(y = x + 2\)
library(geiger)
## Loading required package: ape
## Warning: package 'ape' was built under R version 4.1.2
x = seq(-10, 10, by=1)
f1 = (-10:10)^2 - 2*(-10:10) - 2
f2 = (-10:10) + 2
geiger:::.area.between.curves(x, f1, f2)
## [1] 5
  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.

C = cost n = number of orders x = number of irons in the order nx = 110

Tricky part: at any given time, there are on average x/2 irons in storage.

\(C = \frac{x}{2}∗3.75+8.25∗n\)

\(C=\frac{\frac{110}{n}}{2}∗3.75+8.25∗n\)

\(C=\frac{110}{2n}∗3.75+8.25∗n\)

\(C=55\frac{1}{n}∗3.75+8.25∗n\)

\(C'=-55\frac{1}{n^{-2}}∗3.75+8.25\)

We need to take the derivative and solve for 0.

\(0=-55\frac{1}{n^{2}}∗3.75+8.25\)

\(0=-55\frac{1}{n^{2}}∗3.75+8.25\)

\(\frac{-8.25}{-206.25}=\frac{1}{n^{2}}\)

\(\frac{-206.25}{-8.25}={n^{2}}\)

n = 5

5 * x = 110

x = 22

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

\(\int ln(9x) * x^6\mathrm{d}x\)

\(u = ln(9x)\)

\(dv = x^6\mathrm{d}x\)

\(du = \frac{1}{x}\mathrm{d}x\)

\(v = \frac{x^7}{7}\)

\(\int u\mathrm{d}v = uv - \int v\mathrm{d}u\)

\(\int ln(9x) * x^6\mathrm{d}x = ln(9x)\frac{x^7}{7} - \int \frac{x^7}{7} \frac{1}{x}\mathrm{d}x\)

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

  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.

\(\int^{e^6}_{1} \frac{1}{6x}dx\)

\(\frac{1}{6}ln(x)]^{e^6}_{1}\)

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

\(\frac{6}{6} - 0\)

It is a probability density function