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

\[∫4e^{−7x}dx\]

\[u=−7x\]

\[du=−7dx\]

\[dx=\frac{du}{−7}\]

\[∫4e^u\frac{du}{−7}\]

\[\frac{4}{−7}∫e^udu\]

\[\frac{4}{−7}e^u+C\]

\[\frac{4}{−7}e^{−7x}+C\]

  1. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of dNdt=−3150t4−220 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.

\[\frac{dN}{dt}=−\frac{3150}{t^4}−220\]

\[dN=(−\frac{3150}{t^4}−220)dt\]

\[N=∫−\frac{3150}{t^4}−220dt\]

\[N=∫−\frac{3150}{t^4}dt−∫220dt\]

\[N=−\frac{3150}{3t^3}−220t+C\]

\[N(1)=−\frac{3150}{3(1)^3}−220(1)+C=6530\]

\[C=7800\]

\[N(t)=−\frac{31503}{(t)^3}−220(t)+7800\]

\[N(0)=−\frac{3150}{3(0)^3}−220(0)+7800=7800\]

  1. Find the total area of the red rectangles in the figure below, where the equation of the line is:

\[f ( x ) = 2x  9.\]

\[∫^{8.5}_{4.5}2x−9dx\]

\[=[x^2−9x]|^{8.5}_{4.5}\]

\[=(8.5^2−9(8.5))−(4.5^2−9(4.5))\]

\[=16\]

*** Area rectangles is equal to 16 ***

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

\[y=x^2−2x−2, y=x+2\]

Graph for equation:

curve(x^2 -2*x-2, lwd = 2, xlim=c(-5, 5))
curve(x+2, lwd = 2, xlim=c(-5, 5), add = TRUE)

Intersection: \[x^2 -2x-2 = x+2 x^2 -3x-4 = 0\]

f <- function(a){ a^2 - 3*a - 4 }

root <- polyroot(c(-4, -3, 1)) 

ifelse(Im(root) == 0, Re(root), root)
## [1] -1+0i  4-0i

Area: \[∫^4_{−1}x+2dx−∫^4_{−1}x^2−2x−2dx\]

\[=−[\frac{1}{3}x^3−\frac{3}{2}x^2−4x]|^4_{−1}\]

\[=20.8333\]

  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.

Number of orders per year = n lot size = s Cost = c

\[ns = 110 s = 110 / n\]

Assume half of inventory keep in stocks:

\[c=8.25n+\frac{375}{2n}\]

\[c=8.25n+\frac{206.25}{n}\]

\[c′=8.25−\frac{206.25}{n^2}\]

\[c′=0\]

\[0=8.25−\frac{206.25}{n^2}\]

\[n=5\]

*** Orders per year is 5 ***

*** Lot size is 22 and inventory cost is 78.75.***

  1. Use integration by parts to solve the integral below

\[∫ln(9x)∗x^6dx\]

Choose:

\[u=ln(9x), \frac{dv}{dx}=x^6\]

\[du=\frac{9}{9x}dx=\frac{1}{x}dx\]

\[dv=x^6dx\]

\[v=\frac{1}{7}x^7\]

In equation:

\[∫udv=uv−∫vdu\]

\[=ln(9x)\frac{1}{7}x^7−∫\frac{1}{7}x^7\frac{1}{x}dx\]

\[=ln(9x)\frac{x^7}{7}−\frac{x^7}{49}−C\]

  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.

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

\[∫^{e6}_1f(x)dx\]

\[=∫^{e^6}_1\frac{1}{6x}dx\]

\[=\frac{1}{6}∫^{e^6}_1\frac{1}{x}dx\]

\[=\frac{1}{6}ln(x)|^{e^6}_1\]

\[=\frac{1}{6}[ln(e^6)−ln(1)]\]

\[=\frac{1}{6}[6−0]\]

\[=1\]