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