# Question 1

Use integration by substitution to solve the integral below. $\int 4e^{-7x}dx$

## Solution

$Let\: u = -7x,\: \therefore du = -7 dx \: and \: dx = \frac{du}{-7}$

$\int \frac{4e^{u}du}{-7}$

$\frac{-4}{7}\int e^u du$

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

# Question 2

Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of $$\frac{\text{d}N}{\text{d}t} = -\frac{3150}{t^4} - 220$$ 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 day 1 was 6530 per cubic centimeter.

## Solution

$\frac{\text{d}N}{\text{d}t} = -\frac{3150}{t^4} - 220 \$

$N'(t) = \int (-\frac{3150}{t^4}-220)dt$

$-3150 \int \frac{1}{t^4}dt-220 \int1dt$

$(-3150)(-\frac{1}{3t^3})-220(t)$

$\frac{1050}{t^{3}} - 220t$

$N(t) = \frac{1050}{t^{3}} - 220t + C$

$N(1) = 1050 - 220 + C = 6530$

$C = 6530 - 1050 + 220 = 5700$

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

# Question 3

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

## Solution

$\int_{4.5}^{8.5}(2x - 9)dx$

$2 \int x \ dx - 9 \int 1 \ dx$

$2(\frac{x^2}{2})-9(x)$

$x^2-9x |_{4.5}^{8.5}$

$(8.5^2-9\cdot8.5)-(4.5^2 - 9\cdot4.5)=16$

# Question 4

Find the area of the region bounded by the graphs of the given equations $y = x^2-2x-2, \ \ y = x + 2$

## Solution

f1 <- function(x) {x^2 - 2*x - 2}
f2 <- function(x) {x + 2}
shade <- function(x) {y <- f1(x)
y[x < -1 | x > 4] <- NA
return(y)}

ggplot(data.frame(x=c(-2, 5)), aes(x = x)) +
stat_function(fun = shade, geom = "polygon", fill = "grey", alpha = 0.7) +
stat_function(fun = f1, geom = "line", aes(colour = "f1")) +
stat_function(fun = f2, geom = "line", aes(colour = "f2")) +
theme(legend.position = "none")

$y = x^2-2x-2, \ \ y = x + 2$

$x^2-2x-2 = x + 2$

$x^2-2x-2 - x - 2=0$

$x^2-3x-4=0$

$(x-4)(x+1)=0$

$x=\{4,-1\}$

$\int_{-1}^4((x+2)-(x^2-2x-2))dx$

$\int_{-1}^4(-x^2+3x+4)dx$

$-\int x^2\:dx+3\int x\:dx + 4 \int1\:dx$

$-\frac{x^3}{3}+ \frac{3x^2}{2} + 4x |_{-1}^4$

$(-\frac{4^3}{3}+ \frac{3\cdot4^2}{2} + 4\cdot4) - (-\frac{-1^3}{3}+ \frac{3\cdot-1^2}{2} + 4\cdot-1)=20.8\bar{3}$

# Question 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.

## Solution

Reference:

https://en.wikipedia.org/wiki/Economic_order_quantity

$D=annual\ demand\ quantity=110$

$K=fixed\ cost\ per\ order=8.25$

$h=storage\ cost\ per\ unit=3.75$

$Q=optimal\ order\ quantity$

$Q^* = \sqrt{\frac{2DK}{h}}$

$Q^* = \sqrt{\frac{2\cdot 110\cdot 8.25}{3.75}} = \sqrt{\frac{1815}{3.75}} = \sqrt{484} = 22$

$orders =\frac{110}{22}=5$

# Question 6

Use integration by parts to solve the integral below:

$\int \ln(9x)\cdot x^6 \:dx$

## Solution

$\int fg'=fg-\int f'g$

$f=ln(9x),\ g'=x^6$

$f'=\frac{1}{x},\ g=\frac{x^7}{7}$

$\frac{x^7ln(9x)}{7}-\int \frac{x^6}{7}\:dx$

$\frac{x^7ln(9x)}{7}-\frac{1}{7}\int x^6\:dx$

$\frac{x^7ln(9x)}{7}-\frac{1}{7} \cdot \frac{x^7}{7}$

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

# Question 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.

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

## Solution

$\int_1^{e6} \frac{1}{6x}dx$

$\frac{1}{6}\int \frac{1}{x}dx$

$\frac{1}{6} \cdot ln(x)$

$\frac{ln(x)}{6} |_1^{e^6}$

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

$1-0=1$

“The probability density function is nonnegative everywhere, and its integral over the entire space is equal to 1.” - Wikipedia PDF Definition