Problem 1

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

Solution

Let \(u=-7x\) and \(du = -7dx\). Therefore, \(4dx = -\frac{4}{7}du\).
By \(u\) substitution: \[\int -\frac{4}{7}e^{u}dx\] \[-\frac{4}{7} \int e^{u}dx\] \[-\frac{4}{7} e^{u} + C\]

Substituting back in for \(u\), we get: \[-\frac{4}{7} e^{-7x} + C\]

Problem 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 pf days since treatment began. Find a function \(N(t)\) to estimate the level of contamination if the level after day 1 was \(6,530\) bacteria per cubic centimeter.

Solution

We know that \(\frac{dN}{dt} = N'(t)\) so in order to solve for \(N(t)\), we must take the integral. \[N(t) = \int - \frac{3150}{t^4} - 220 dt\] \[N(t) = \frac{3150}{3t^3} - 220t + C\] \[N(t) = \frac{1050}{t^3} - 220t + C\]

In order to solve for \(C\), we will use the fact that \(N(1) = 6530\). \[N(1) = 6530= \frac{1050}{1^3} - 220(1) + C\] \[6530= 1050 - 220 + C\] \[C = 5700\]

Therefore, we know that: \[N(t) = \frac{1050}{t^3} - 220t+ 5700\]

Problem 3

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

Solution

\[\int_{4.5}^{8.5}x 2x - 9 dx\] \[= x^2 - 9x \Big|_{4.5}^{8.5}\] \[= (8.5^2 - 9(8.5)) - (4.5^2 - 9(4.5))\] \[= (72.25 - 76.5) - (20.25 - 40.5)\] \[= -4.25 - -20.25\] \[= 16\]

Problem 4

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

Solution

First, we will see where the two functions intersect by setting them equal to each other: \[x^2 - 2x - 4 = x + 2\] \[x^2 - 3x -4 = 0\] \[(x+1)(x-4)= 0\] \[x = -1, x = 4\]

To find the area in between the two curves, we can use the following: \[A = \int_{a}^{b} (upper function) - (lower function) dx\]

We can see from plotting the curves that our upper function is \(y = x + 2\) and our lower function is \(y = x^2 - 2x - 2\).

y1 <- function(x){x^2 - (2*x) - 2}
y2 <- function(x){x + 2}

curve(y1, from = -1, to = 4)
curve(y2, from = -1, to = 4, add = TRUE)

We know our bounds are \([-1,4]\), so we can plug this information into our formula and solve: \[\int_{-1}^{4} (x + 2) - (x^2 - 2x - 2) dx\] \[\int_{-1}^{4} - x^2 + 3x + 4 dx\]

Integrating and plugging in our bounds we get:

fun <- function(x) (3*x) + 4 - x^2
integrate(fun,-1,4)
## 20.83333 with absolute error < 2.3e-13

Problem 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

First, we will need to define a function for the inventory cost that takes into account the storage costs and order costs.

In order to minimize this function, we will take its derivative: \[f(n) = 3.75*\frac{n}{2} + 8.25 * \frac{110}{n}\] \[f(n) = 1.875n + \frac{907.5}{n}\] \[f'(n) = 1.875 - \frac{907.5}{n^2}\] \[0 = 1.875 - \frac{907.5}{n^2}\] \[\frac{907.5}{n^2} = 1.875 \] \[907.5 = 1.875(n^2) \] \[484 = n^2 \] \[n = 22\]

Subbing this back into our equations, we know that the number of orders per year to minimize inventory costs is: \(\frac{110}{n} = \frac{110}{22} = 5\)

Problem 6

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

Solution

Let \(u=ln(9x)\), \(du = \frac{1}{x}\), \(v = \frac{1}{7}x^7\), and \(dv = x^6\). \[\int udv = uv- \int vdu\] \[= ln(9x) \frac{1}{7}x^7 - \int \frac{1}{x} * \frac{1}{7}x^7 dx\] \[= \frac{1}{7}ln(9x)x^7 - \frac{1}{7}\int \frac{x^7}{x} dx\] \[= \frac{1}{7}ln(9x)x^7 - \frac{1}{7}\int x^6 dx\] \[= \frac{1}{7}ln(9x)x^7 - \frac{1}{7} * \frac{1}{7}x^7 + C\] \[= \frac{1}{7}ln(9x)x^7 - \frac{1}{49}x^7 + C\]

Problem 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

To satisfy the conditions of a probability density function, the total area within the bounds must be 1. \[\int_{1}^{e^6} \frac{1}{6x} dx \] \[\frac{1}{6} \int_{1}^{e^6} \frac{1}{x}\] \[\frac{1}{6} ln(x) |_{1}^{e^6}\]

\[\frac{1}{6} (6) - \frac{1}{6} (0) \] \[= 1 \]

Yes, this is a probability density function.