1.

Use integration by substitution to solve the integral below.

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

a. Sub in U for -7x

\(u = -7x\)

\(du = -7dx\)

\(dx = \frac{du}{-7}\)

b. Re-write as:

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

c. Pull out dx

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

d. Lastly integrate remaining problem and replace u

\[\frac{-4}{7}e^{u}+C \rightarrow \frac{-4}{7}e^{-7x}+C\]

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.

a.

\[\int \frac{-3150}{t^{4}}-220\]

b. pull t^4 into the numerator, making it negative

\[\int (-3150t^{-4}-220)dt\]

c. pull out -3150

\[-3150 \int t^{-4} - \int -220 dt\]

d. take the integral

\[-3150 \frac{-1}{3}t^{-3} - 220t + C\]

e. simplify

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

f. Solve for N(t)=6530 and t=1

Find a function N( t ) to estimate the level of contamination if the level after 1 day was 6530 bacteria per cubic centimeter.

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

\[6530 = 1050-220+C\]

\[6530 = 830+C\]

\[C = 5700\]

g. Add C back in

\[N(t) = \frac{1050}{t^3}-220t+5700\]

3.

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

a. The values are between 4.5 and 8.5 on the x-axis, thus I will integrate between these two values.

\[\int_{4.5}^{8.5}(2x -9)dx \rightarrow (x^2 -9)|_{4.5}^{8.5}\]

b. Feed in the limits and take the difference

\[8.5^2 - 9(8.5) - (4.5^2 -9(4.5)) = 16\]

The total area under f(x) = 2x-9 is equal to 16

4. Going to try a function for this step using the integrate() function.

upper <- 4
lower <- -1
e1 <- function(x) {x + 2}
e2 <- function(x) {x^2 - 2*x -2}

area_e1 <- integrate(e1, lower, upper)
area_e2 <- integrate(e2, lower, upper)
solution <- area_e1$value - area_e2$value
print(solution)
## [1] 20.83333

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.

ex <- 110
fixed <- 8.25
store <- 3.75

lot_size <- sqrt((2*ex*fixed)/store)
print(lot_size)
## [1] 22
orders_py <- ex/lot_size
print(orders_py)
## [1] 5

6.

Use integration by parts to solve the integral below.

\[\int ln(9x) * x^6\] Let: \[ u=ln(9x)\\ du=\frac{1}{x}dx\\ dv=x^{6}\\ v=\frac{x^7}{7} \]

Substitute in values into the substitution by parts formula:

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

7.

Determine whether f ( x ) is a probability density function on the interval [1, e6] . If not, determine the value of the definite integral.

The sum of the definite integral f(x) is 1 so f(x) is a probability density function on the interval [1, e6].

\[ f(x) = \frac{1}{6x} |_1^{e^6}\\ \int_1^{e^6} f(x) = \frac{ln(e^6)}{6} - \frac{ln(1)}{6} = \frac{ln(e^6)}{6} - \frac{ln(1)}{6} = 1-0 = 0 \]