1.

Use integration by substitution to solve the integral below.

\[\int 4e^{-7x} dx\]

Let \(\ u = -7x\)

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

\[\frac{4}{-7}[e^{-7x}] + c\]

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.

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

\[\ dN =(\frac{-3150}{t^4} - 220)dt\]

\[\ N=∫\frac{-3150}{t^4}−220dt\]

\[\ N=∫\frac{-3150}{t^4}dt−∫220dt\]

\[\ N=\frac{−3150}{3t^3}−220t+C\]

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

\[\ C=7800\]

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

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

3.

Find the total area of the red rectangles in the figure below, where the equation of the line is f ( x ) = 2x  9. Area = Find the area of the region bounded by the graphs of the given equations.

\[\int_{4.5}^{8.5} 2x-9dx\]

\[\ =[x2−9x]|_{4.5}^{8.5}\]

\[\ =(8.52−9(8.5))−(4.5^2−9(4.5))\]

\[\ =16\]

\[\ y=x^2−2x−2, y=x+2\]

4.

Enter your answer below.

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

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


# zeros of f 
root <- polyroot(c(-4, -3, 1)) 
ifelse(Im(root) == 0, Re(root), root)
## [1] -1+0i  4-0i

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

\[\ =−[13x^3−32x^2−4x]|_{−1}^{4}\]

\[\ =20.8333\]

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.

Number of orders per year = n

Lot size = s

Cost = c

n*s = 110

s = 110 / n

Half of inventory keep in stocks:

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

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

\[\ c′=8.25−\frac{206.25}{n^2}\]

\[\ c′=0\]

\[\ 0=8.25−\frac{206.25}{n^2}\]

\[\ n=5\]

Number of order per year is 5

Lot size is 22 Inventory cost is 78.75

6.

Use integration by parts to solve the integral below. \[\int ln(9x)∗x^6dx\]

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:

\[\int udv=uv−\int vdu\]

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

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

7.

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

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

\[\int_{1}^{e^6}f(x)dx\]

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

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

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

\[\ =\frac{1}{6}[ln(e^6)−ln(1)]\]

\[\ =\frac{1}{6}[6−0]\]

\[\ =1\]