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

Solution

Note: \(\int{e^u du} = e^u\) \[ \int{4e^{-7x}dx}\] Lets v = -7x dv = dx substitute this below: \[ \int{4e^{-7x}dx} = 4\int{e^{v}dv} = {4}e^{v} + C = -\frac{4}{7}e^{-7x} + C \] 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.

Solution

Given,

t = 1

N(1) = 6530

\[\frac{dN}{dt}= -\frac{3150}{t^4} - 220 = N(t) = -\int\frac{3150}{t^4} dt - 200 \int dt+ C\]

\[N(t) = -3150*t^{-4}- 200 \int dt+ C = -\frac{3150} {-3} t ^ {-3} - 200 (t) + C \] Substitue value of t and N(t) :

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

\[C = 6530 - 1050 + 220\] \[C = 5700 \]

Substitute C value:

\[N(t) = 1050 t^{-3} - 200 t + 5700\]

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

Figure 1

Solution

Above plot shows limit (lower = 4.5, upper = 8.5)

area_fun <- function(x) 
  {
  return(2 * x - 9)
}

value <- integrate(area_fun, lower = 4.5, upper = 8.5)$value

print(value)
## [1] 16

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

Solution

value of x

\[x ^ 2 - 2x -2 = x + 2\]

\[x ^ 2 - 3x -4 = 0\]

\[x ^ 2 - 4x + x -4 = 0\]

\[x(x - 4) + 1(x - 4) = 0\] \[(x - 4) * (x + 1) = 0\] x = (-1, 4)

Create 2 functions for 2 equations.

graph_1 <- function(x){
  return(x^2 - 2*x -2)
}

graph_2 <- function(x){
  return(x +2)
}

Plot graphs.

plot(graph_1, -5, 5, col = 'purple', ylab = "Y")
plot(graph_2, -5, 5, col = '#05bbaa', add = TRUE)

Area of intersection

func_intersect <- function(x)
  {
    return((x+2) - (x^2 - 2*x -2))
}

area <- integrate(func_intersect, lower = -1, upper = 4)$value
print(area)
## [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.

Solution

Suppose half of inventory keeps stock, let x = size of iron

storage cost = 3.75 * x/2

order cost = 8.25 * 110/x

Total Cost = 1.875x + 907.5/x

First order derivate 1.875 - 907.5/\(x^2\) = 0

\[1.875 * x ^ 2 - 907.5 = 0\] \[1.875 * x ^ 2 = 907.5\] \[x ^ 2 = \frac {907.5}{1.875}\] \[x = \sqrt \frac {907.5}{1.875}\]

x <- sqrt(907.5/1.875)
print(x)
## [1] 22
times <- 110/x
print(times)
## [1] 5

Each year iron can order 5 times size of 22.

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

Solution

Integration by parts formula:

u = ln(9x)

du = \(\frac {1}{x} dx\)

v = x^6

dv = \(\frac{1}{7}x^7\)

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

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

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

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