Exercise 1

Use integration by substitution below \[ \int 4 e^{-7x} dx \]

Solution

\[ \text{Let u = -7x}, \text{and then} \ -\frac{1}{7} du = dx \\ \int 4 e^{7x} dx \Rightarrow 4 \int e^u du \\ \Rightarrow -\frac{4}{7} e^{-7x} + C \]

Exercise 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.

Solution

\[ \int \frac{d}{dt}N = \int -\frac{3150}{t^4} - 220 dt \\ \Rightarrow \frac{-3150 t^{-3}}{-3} - 220 t + C \\ \Rightarrow N(t) = \frac{1050}{t^3} -220t + C \] After 1 day there was 6530 bacteria per cm^3

\[ N(t) = 6530 \\ 6530 = \frac{1050}{1^3} - 220(1) + C \\ C = 5700 \]

Exercise 3

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

domain looks to be from 4.5 to 8.5

Solution

We have,

\[ \int_{4.5}^{8.5} 2x - 9 dx \Rightarrow x^2-9x |_{4.5}^{8.5} = (8.5^2 -9(8.5)) - (4.5^2 - 9 (4.5)) = 16 = \text{Area under the line} \]

Exercise 4

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

Solution

Let \(f(x) = x^2-2x-2\) and \(g(x)=x+2\),

\[ \int_{-1}^{4} f(x) - g(x) dx = \int_{1}^4 x^2-2x-2 - (x +2) dx = \int_{1}^4 x^2 -3x - 4 dx \\ \Rightarrow \frac{x^3}{3} - \frac{x^2}{2} -4x |_{-1}^4 = [\frac{4^3}{3} - \frac{4^2}{2} -4(4)] - [ \frac{-1^3}{3} - \frac{-1^2}{2} -4(-1)] \\ \Rightarrow -20.83 \]

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

Exercise 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 define the our variables

  • TC = Total Cost
  • Q = Quantity
  • HC = Holding Cost = 3.75
  • OC = Ordering Cost = 8.25
  • D = Demand = 110

We have the following function

\[ TC(Q) = \frac{D*OC}{Q} + \frac{HC*Q}{2} \] To minimize, we will need to the derivative and solve for Q.

\[ \frac{d}{dQ} TC = - \frac{D*OC}{Q^2} + \frac{HC}{2} = 0 \\ \Rightarrow \frac{HC}{2} = \frac{D*OC}{Q^2} \\ \Rightarrow Q^2 = \frac{2*D*OC}{HC} \\ \Rightarrow Q = \sqrt{\frac{2*D*OC}{HC}} \\ \Rightarrow Q = \sqrt{\frac{2*110*8.25}{3.75}} = 22 \]

Which gives us the formula for the Economic Ordering Quantity in Economics.

Order per year = Demand / Q

\[ Orders = 110 / 22 = 5 \]

Exercise 6

Use Integration by Parts to solve the integral below \[ \int ln(9x)x^6 dx \]

Solution

\[ \begin{align} u = ln(9x), &\ dv = x^6 dx \\ du = \frac{1}{x}, &\ v = \frac{1}{7}x^7 \end{align} \] \[ ln(9x) \frac{1}{7}x^7 - \int \frac{1}{7}x^7 \frac{1}{x} \\ \Rightarrow \frac{x^7 ln(9x)}{7} - \frac{1}{7} \int x^6 dx \\ \Rightarrow \frac{x^7 ln(9x)}{7} - \frac{1}{7}(\frac{x^7}{7} + C) \\ \Rightarrow \frac{x^7 ln(9x)}{7} - \frac{x^7}{49} + C) \]

Exercise 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

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

So, since f(x) = 1 on the interval, this is a pdf.