Exercise #1

Use integration by substitution to solve the integral below.

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

Let:

\[ \begin {align} u &= -7x \\ du &= -7 dx \\ dx &= \frac{du}{-7} \end{align} \]

Now, let’s substitute in the original integral:

\[ \begin{align} \int 4 e^{-7x} dx &= \int 4 e^u \frac {du}{-7} \\ \frac{-4}{7} e^u + C &= \frac{-4}{7} e^{-7x} + C\\ \boxed{\int 4 e^{-7x} dx=\frac{-4}{7} e^{-7x} + C} \end{align} \]

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.

To find the function \(N(t)\), we need to solve:

$ - 220 dt $

\[ \begin{align} \int \frac{-3150}{t^4} - 220 dt &= \int \frac{-3150}{t^4} dt + \int -220 dt \\ -3150 \int t^{-4} dt -220 \int dt &= -3150 (\frac {t^{-3}}{-3}) -220 t + C \\ int \frac{-3150}{t^4} - 220 dt &=\frac{1050}{t^3} - 220t + C\\ \boxed {N(t) = \frac{1050}{t^3} - 220t + C} \end{align} \]

Exercise #3:

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

To find the total area of the red rectangles in the figure given the equation of the line, we need to calculate the integral of the given equation on the interval [4.5, 8.5]

\[ \begin {align} \int _{4.5}^{8.5} 2x-9 dx &= x^2 - 9x \bracevert _{4.5} ^{8.5}\\ [4.5^2 - (9*4.5)] - [8.5^2 - (9*8.5)] &= 16 \end {align} \]

Let’s check if the area is \(16 units^2\)

# Define the function
f1 <- function(x) {
  return((2*x) - 9)
}

# Calculate the definite integral
Area1 <- integrate(f1, lower = 4.5, upper = 8.5)

# Print the result
print(Area1)
## 16 with absolute error < 1.8e-13

Exercise #4:

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

First, we need to find the boundaries of the integral; we find the points of intersection between the two lines:

\[ \begin {align} x^2 -2x -2 = x+2 \\ \Rightarrow x^2 - 3x -4 = 0 \end{align} \]

Using the quadratic equation : \(x = \frac{-b \pm \sqrt {b^2 -4ac}}{2a}\) where \(a=1\) , \(b=-3\) and \(c=-4\)

\[ \begin {align} x &= \frac{3 \pm \sqrt {(-3)^2 -4(1)(-4)}}{2(1)}\\ x &= \frac{3 \pm \sqrt{9+16}}{2}\\ x &= \frac{3 \pm 5}{2} \\ x &= -1 \\ \text{or}\\ x&= 4 \end{align} \]

The interval we are going to integrate the difference between the given equations is $[-1,4].

Now, let’s evaluate both equations in numbers that are in the interval \(x=2\) and \(x=3\) to determine which one is above the other.

For \(x=2\)

\[ \begin{align} y &= (2)^2 -(2)(2) -2\\ y &= 4-4-2 = -2 \\ \text{and}\\ y &= (2) + 2 = 4 \end{align} \]

For \(x=3\)

\[ \begin{align} y &= (3)^2 -(2)(3) -2\\ y &= 9-6 -2 = 1 \\ \text{and}\\ y &= (3) + 2 = 5 \end{align} \] So, the equation \(x+2\) is above the equation \(x^2 -2x -2\)

\(\Rightarrow\) The equation we need to integrate is:

\[ \begin{align} x+2 - (x^2 - 2x - 2) &= x + 2 - x^2 + 2x + 2 \\ &= -x^2 + 3x +4 \end{align} \]

Let’s find the area:

\[ \begin{align} \int_{-1}^{4} (-x^2 + 3x +4 )dx & = \int _{-1}^{4} -x^2 dx + \int _{-1}^{4} 3x dx + \int _{-1}^{4} 4dx \\ &= [-\frac {1}{3} x^3]_{-1}^{4} + [\frac{3}{2} x^2]_{-1}^{4} + [4x]_{-1}^{4} \\ &= [\frac{-1}{3} (64+1)] + [\frac{3}{2}(16-1)] +[4(4+1)] \\ &= \frac{-1}{3} (65) + \frac{3}{2} (15) + 20 \\ &= \frac{-65(2) +3(15)(3)}{6} +20 \\ &= \frac{5}{6} + 20 = 20.8 \overline{3} \end{align} \]

Let’s check our answer;

f2 <- function(x) {
  return(-(x^2) + (3 * x) + 4 )
}
Area2 <- integrate (f2, lower=-1, upper=4 )
print(Area2)
## 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.

“To minimize inventory costs for the beauty supply store, we need to find the optimal lot size (number of flat irons per order) and the number of orders per year. The total inventory cost consists of two components: the cost of holding inventory (storage cost) and the cost of placing orders.

Let’s denote the following variables:

  • D: demand for flat irons in a year (110 flat irons)

  • S: cost of storing one flat iron for one year ($3.75)

  • K: fixed cost for each order ($8.25)

  • Q: lot size (number of flat irons per order)

  • N: number of orders per year

The total inventory cost (TC) can be expressed as: \(TC=DS+ \frac{D}{Q} K\)

To minimize TC, we need to find the values of Q and N that minimize this function.

Step 1: Express the Total Cost Function

\(TC(Q, N) = DS + \frac{D}{Q} K\)

Step 2: Partial Derivatives

Compute the partial derivatives of TC with respect to Q and N:

\(\frac{dTC}{dQ} = \frac{-DK}{Q^2}\) \(\frac{dTC}{dN} = \frac{-DK}{Q}\)

Step 3: Set Partial Derivatives Equal to Zero

To find the critical points, set the partial derivatives equal to zero:

\(\frac{-DK}{Q^2} = 0\)

\(\frac{-DK}{Q}=0\)

Step 4: Solve for Q and N

Solve the equations to find the values of Q and N.

From the first equation, we get \(Q=\sqrt{\frac{DK}{S}}\)

From the second equation, we find $ N= $

**Step 5: Evaluate $TC(Q, N) $

Evaluate the total cost function at the optimal values of Q and N to find the minimum inventory cost.

Let’s calculate the optimal values:

\(Q= \sqrt{\frac{(110)(8.25)}{3.75}} \approx 12.81\)

\(N = \frac{110}{12.81} \approx 8.59\)

Since we can’t have fractional values for the lot size and number of orders, we can round Q up to 13 and N up to 9.

So, the optimal lot size is 13 flat irons per order, and the number of orders per year is 9.

Note: the solution to this problem, i took it from ChatGPT, because I couldn’t understand it

Exercise #6

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

For this problem, we are going to use the integration by parts rule: \(\int uv dx = u \int v dx - \int u' (\int v dx) dx\)

Let:

\[ \begin{align} u&=ln(9x)\\ v&=x^6 \\ u'&= \frac{1}{9x} \times 9 = \frac{1}{x} \\ \int x^6 dx &= \frac{1}{7} x^7 \\ \end{align} \]

Let’s substitute:

\[ \begin {align} \int ln(9x) \cdot x^6 &= ln(9x)(\frac{1}{7} x^7) - \int \frac{1}{x} (\frac{1}{7} x^7) dx\\ &= \frac{1}{7} x^7 ln(9x) - \int \frac{x^6}{7} dx \\ &= \frac{1}{7} x^7 ln(9x) - \frac{1}{7} (\frac{x^7}{7}) + C\\ &= \frac {x^7 ln(9x)}{7} - \frac{x^7}{49} + C \\ &= \frac{7 x^7 ln(9x) - x^7}{ 49} + C\\ &= \frac{x^7 (7 ln(9x) -1)}{49} + C\\ \boxed{\int ln(9x) \cdot x^6 =\frac{x^7 (7 ln(9x) -1)}{49} + C} \end{align} \]

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

To determine if the function is a probability density function, we need to check if:

  1. \(f(X) > 0\) in the interval \([1, e^6]\)

  2. \(\int_{1}^{e^6} f(x) dx = 1\)

Let’s do this:

\(x>0\) because \(1<x<e^6\) \(\Rightarrow 6x >0\) so \(\frac{1}{6x} > 0 \rightarrow f(x)>0\)

\[ \begin{align} \int_{1}^{e^6} f(x) dx &= \int_{1}^{e^6} \frac{1}{6x} dx \\ \frac{1}{6} \int_{1}^{e^6} \frac{1}{x} dx &= \frac{1}{6} ln(x) \bracevert _1 ^{e^6} \\ &= \frac{1}{6} [ln(e^6) - ln(1) ] \\ &= \frac{1}{6} [6-0] = \frac{1}{6} (6) = 1 \end{align} \] Check our answer:

f3 <- function(x) {
  return(1/(6*x))
  }
Area3 <- integrate (f3, lower=1, upper=exp(6) )
print(Area3)
## 1 with absolute error < 9.3e-05

The function \(f ( x )= \frac{1}{6x}\) is a probability density function on the interval \([1, e^6]\)