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

\(\int 4e^{-7x}\)

Set \(u = -7x\). The derivative of \(-7x\) is \(-7\), so \(du = -7dx\). Solving for \(dx\), we get \(dx = -\frac{1}{7}du\). Substituting \(u\) and \(dx\), we get:

\(\int -\frac{4}{7}e^{u}du\), or \(-\frac{4}{7}\int e^{u}du\). The integral of \(e^u\) is \(e^u\), so:

\(-\frac{4}{7}\int e^{u}du = -\frac{4}{7}e^u\)

Substituting \(u\) back into the result and adding the constant, the solution is:

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


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.

Start by taking the integral of boths sides of \(\frac{dN}{dt} = -\frac{3150}{t^4}-220\):

\(\int \frac{dN}{dt} = \int -3150t^{-4} - 220\); results in:

\(N(t) = -3150(-\frac{1}{3}t^{-3}) - 220t + C\)

Then, substitute \(t = 1\) and \(N(t) = 6530\) to compute \(C\):

\(6530 = -3150(-\frac{1}{3}(1^{-3}) - 220(1) + C = \frac{3150}{3} - 220 + C = 1050 - 220 + C\) or

\(6530 = 830 + C\). Therefore, \(C = 6530 - 830 = 5700\). Substituting C back into the \(N(t)\) equation and simplifying:

\(N(t) = -3150(-\frac{1}{3}t^{-3}) - 220t + 5700\)

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


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


This problem can be solved using Riemann sums with the Left Hand Rule, which would evaluate the area as

\(f(5) * 1 + f(6) * 1 + f(7) * 1 + f(8) * 1\):

\((5 * 2 - 9) + (6 * 2 - 9) + (7 * 2 - 9) + (8 * 2 - 9) = 1 + 3 + 5 + 7 = 16\)

\(Area = 16\)


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

\(y = x^2 - 2x - 2, y = x + 2\)

# create functions f and g; f is the 1st function, g is the 2nd function
f <- function(x) {x^2-2*x-2}
g <- function(x) {x+2}
# plot to visualize
curve(f, -2, 7)
curve(g, -2, 7, add=TRUE)

# confirm points of intersection
# intersection where x below zero is -1:
rt <- uniroot(function(x)  f(x) - g(x)  , c(-5000,-0.01), tol=1e-8) 
rt$root
## [1] -1
# intersection where x above zero is 4:
rt <- uniroot(function(x)  f(x) - g(x)  , c(0.0001, 5000), tol=1e-8) 
rt$root
## [1] 4
# area between curves is g minus f for x = -1 to 4
# integrate g
g_area <- integrate(g, lower = -1, upper = 4)
# integrate f
f_area <- integrate(f, lower = -1, upper = 4)
# compute difference between g_area and f_area
g_area$value - f_area$value
## [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.

Lot size = flat irons per order = \(s\)
Annual ordering costs (o) = fixed cost/order \(* \frac{110}{s} = 8.25 * \frac{110}{s} = \frac{907.5}{s}\)
Annual inventory costs (i) = storage cost/flat iron \(* \frac{s}{2} = 3.75 * \frac{s}{2} = 1.875s\)
Total Cost (c) = Annual ordering costs (o) + Annual inventory costs (i)

\(c = \frac{907.5}{s} + 1.875s = 907.5s^{-1} + 1.875s\)

\(c' = -907.5s^{-2} + 1.875\); set \(c' = 0\) and solve for \(s\):

\(1.875 - \frac{907.5}{s^2} = 0\) or \(s^2 = \frac{907.5}{1.875}\) or \(s = \sqrt{484} = 22\).

To confirm that 22 is a minimum, we find \(c''\) and confirm that it’s greater than zero for \(s = 22\):

\(c'' = \frac{1815}{s^3}\), when \(s = 22\) then \(c'' > 0\), so it’s a minimum.

So the lot size which minimizes inventory costs is 22, which means that the number of orders per year is 5: \(\frac{110}{22} = 5\)


6. Use integration by parts to solve the integral below.

\(\int \ln(9x)x^6dx\)

Integration by parts formula: \(\int u dv = uv - \int v du\)

I set \(u = \ln(9x)\) and \(dv = x^6 dx\). Accordingly, \(du = \frac{1}{x} dx\) and \(v = \int x^6 dx = \frac{x^7}{7}\). Plugging \(u\), \(v\), and \(du\) into the integration by parts formula:

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


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

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

If the sum of the definite integral for the function on interval \(\big[1, e^6\big]\) is \(1\), then it is a probability density function, because the sum of all possible probabilities is always \(1\) in a probability density function.

Compute \(\int_{1}^{e^6} \frac{1}{6x} dx\).

\(\int \frac{1}{6x} dx = \frac{\ln(x)}{6}\), so \(\int_{1}^{e^6} \frac{1}{6x} dx = \frac{\ln(e^6)}{6} - \frac{\ln(1)}{6} = 1 - 0 = \boldsymbol{1}\).

So, \(f(x)\) is a probability density function because the value of the definite integral is 1.