library(ggplot2)

## Warning: package 'ggplot2' was built under R version 3.5.3

1, Use integration by substitution to solve the integral $\int 4e^{-7x}dx$

Solution: Let $u = -7x$. Then $du = -7 dx \ \to \ dx = \frac{du}{-7}$.

Our integral is now $\int \frac{4e^{u}du}{-7}$. Taking out the constants: $\frac{4}{-7}\int e^u du$.

Replacing $u$ with our original substitution: $\frac{-4}{7}e^{-7x} + C$.

2, Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of

\[ \frac{\text{d}N}{\text{d}t} = -\frac{3150}{t^4} - 220 \ \to \ \text{d}N = (-\frac{3150}{t^4}-220)\text{d}t \]

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:

To find $N$, we can take the antiderivative, i.e. the integral.

$N = \int (-\frac{3150}{t^4}-220)\text{d}t \ \ = \int -3150(t^{-4}) \text{d}t - \int 220\text{d}t$

Using the power rule for integration: $N = \frac{-3150}{-3}(t^{-3}) - 220t + C$.

for $N(1) = 6530$:

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

$C = 6530 - 1050 + 220 = 5700$.

$N(t) = -1050(t^{-3}) - 200(t) + 5700$.

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

Solution:

The equation is given as $2x - 9$, and the ends of the rectangles look to be 4.5 and 8.5. Since we’re looking for the area, we can integrate this function over these boundaries.

\[ \int_{4.5}^{8.5}(2x - 9)dx \]

Using the power rule for integration:

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

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

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

f1 <- function(x) {x^2 - 2*x - 2}
f2 <- function(x) {x + 2}
funcShaded <- function(x) {
    y <- f1(x)
    y[x < -1 | x > 4] <- NA
    return(y)
}

ggplot(data.frame(x=c(-1, 4)), aes(x = x)) +
  stat_function(fun = funcShaded, geom = "polygon", fill = "grey", alpha = 0.7) +
  stat_function(fun = f1, geom = "line", aes(colour = "f1")) +
  stat_function(fun = f2, geom = "line", aes(colour = "f2"))

\[ A = \int_{-1}^{4} x + 2 \ dx \ - \ \int_{-1}^{4}x^2 - 2x - 2 dx \\ A = \Big[\frac{1}{2}x^2 + 2x\Big]_{-1}^{4} - \Big[\frac{1}{3}x^3 - x^2 - 2x\Big]_{-1}^{4} \\ = -\frac{1}{3}x^3 + \frac{3}{2}x^2 + 4x\Big|_{-1}^{4} \\ \]

$\approx$ 20.8333333

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.

Total Cost = purchase cost or production cost + ordering cost + holding cost

Where:

Purchase cost: This is the variable cost of goods: purchase unit price ?? annual demand quantity. This is P * D

Ordering cost: This is the cost of placing orders: each order has a fixed cost K, and we need to order D/Q times per year. This is K * D/Q

Holding cost: the average quantity in stock (between fully replenished and empty) is Q/2, so this cost is h * Q/2

$TC = PD + \frac{DK}{Q} + \frac{hQ}{2}$. $\frac{\text{d}}{\text{d}Q} = -\frac{DK}{Q^2} + \frac{h}{2}$. #We next set this equal to zero, and solve for $Q$ in order to find the function minimum. $-\frac{DK}{Q^2} + \frac{h}{2} = 0$. $Q^{*2} = \frac{2DK}{h} \ \to \ Q^* = \sqrt{\frac{2DK}{h}}$ We are given these variables, so we just need to plug them into the formula. $D = 110$. $K = 8.25$. $h = 3.75$. \[ Q^* = \sqrt{\frac{2\cdot 110\cdot 8.25}{3.75}} = \sqrt{\frac{1815}{3.75}} = \sqrt{484} = 22. \] #We found 22 to be the lot size per order. #We are given that the store expects to sell $n = 110$ flat irons. If there are $x$ number of irons in each order, our equation is $22\cdot x = 110 \ \to \ x = \frac{110}{22} = 5$. #So to minimize inventory costs, the store should make 5 orders of 22 irons per year.

6, Use integration by parts to solve the integral below

\[ \int \ln(9x)\cdot x^6 \ dx \]

Solution:

The formula for integration by parts is:

\[ \int u \ dv = uv - \int v \ du \] Let $u = \ln(9x)$. Using the chain rule $du = \frac{1}{9x} \cdot 9 \ dx = \frac{1}{x} \ dx$. Let $dv = x^6$, then $v = \int x^6 = \frac{x^7}{7}$. Plugging this into the formula: \[ \ln(9x)\cdot \frac{x^7}{7} - \int \frac{x^7}{7}\cdot \frac{1}{x} \ dx \] We can pull out the constant: \[ \ln(9x)\cdot \frac{x^7}{7} - \frac{1}{7} \int \frac{x^7}{x} \ dx = \ln(9x)\cdot \frac{x^7}{7} - \frac{1}{7} \int x^6 \ dx \] Using the power rule for integration: \[ \ln(9x)\cdot \frac{x^7}{7} - \frac{1}{7} \Big(\frac{x^7}{7}\Big) + C \] \[ = \ln(9x)\cdot \frac{x^7}{7} - \frac{x^7}{49} + C \]

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.

Solution:

\[ \int_{1}^{e^6} \frac{1}{6x} \ dx = \frac{1}{6} \int_{1}^{e^6}\frac{1}{x} \ dx \\ = \frac{1}{6}[\ln(x)]_{1}^{e^6} \\ = \frac{\ln(e^6) - \ln(1)}{6} = \frac{6\cdot \ln(e) - 0}{6} = \frac{6}{6} = 1 \]

Untitled

1, Use integration by substitution to solve the integral \(\int 4e^{-7x}dx\)

Solution: Let \(u = -7x\). Then \(du = -7 dx \ \to \ dx = \frac{du}{-7}\).

Our integral is now \(\int \frac{4e^{u}du}{-7}\). Taking out the constants: \(\frac{4}{-7}\int e^u du\).

Replacing \(u\) with our original substitution: \(\frac{-4}{7}e^{-7x} + C\).

2, Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of

\[ \frac{\text{d}N}{\text{d}t} = -\frac{3150}{t^4} - 220 \ \to \ \text{d}N = (-\frac{3150}{t^4}-220)\text{d}t \]

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:

To find \(N\), we can take the antiderivative, i.e. the integral.

\(N = \int (-\frac{3150}{t^4}-220)\text{d}t \ \ = \int -3150(t^{-4}) \text{d}t - \int 220\text{d}t\)

Using the power rule for integration: \(N = \frac{-3150}{-3}(t^{-3}) - 220t + C\).

for \(N(1) = 6530\):

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

\(C = 6530 - 1050 + 220 = 5700\).

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

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

Solution:

The equation is given as \(2x - 9\), and the ends of the rectangles look to be 4.5 and 8.5. Since we’re looking for the area, we can integrate this function over these boundaries.

\[ \int_{4.5}^{8.5}(2x - 9)dx \]

Using the power rule for integration:

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

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

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

\[ A = \int_{-1}^{4} x + 2 \ dx \ - \ \int_{-1}^{4}x^2 - 2x - 2 dx \\ A = \Big[\frac{1}{2}x^2 + 2x\Big]_{-1}^{4} - \Big[\frac{1}{3}x^3 - x^2 - 2x\Big]_{-1}^{4} \\ = -\frac{1}{3}x^3 + \frac{3}{2}x^2 + 4x\Big|_{-1}^{4} \\ \]

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.

Total Cost = purchase cost or production cost + ordering cost + holding cost

Where:

Purchase cost: This is the variable cost of goods: purchase unit price ?? annual demand quantity. This is P * D

Ordering cost: This is the cost of placing orders: each order has a fixed cost K, and we need to order D/Q times per year. This is K * D/Q

Holding cost: the average quantity in stock (between fully replenished and empty) is Q/2, so this cost is h * Q/2

6, Use integration by parts to solve the integral below

\[ \int \ln(9x)\cdot x^6 \ dx \]

Solution:

The formula for integration by parts is:

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.

Solution:

\[ \int_{1}^{e^6} \frac{1}{6x} \ dx = \frac{1}{6} \int_{1}^{e^6}\frac{1}{x} \ dx \\ = \frac{1}{6}[\ln(x)]_{1}^{e^6} \\ = \frac{\ln(e^6) - \ln(1)}{6} = \frac{6\cdot \ln(e) - 0}{6} = \frac{6}{6} = 1 \]