1: Solve by substitution

To solve the integral

\[ \int 4e^{-7x} \, dx \]

using substitution, let’s perform the substitution \(u = -7x\). Then, \(du/dx = -7\), which implies \(dx = du / -7\).

Now, substituting \(u = -7x\) and \(dx = du / -7\) into the integral, we get:

\[ \int 4e^u \left(\frac{du}{-7}\right) \]

Simplifying, we have:

\[ -\frac{4}{7} \int e^u \, du \]

The integral of \(e^u\) with respect to \(u\) is simply \(e^u\), so:

\[ -\frac{4}{7} e^u + C \]

Now, we need to back substitute for \(u\):

\[ -\frac{4}{7} e^{-7x} + C \]

So, the antiderivative of \(4e^{-7x}\) with respect to \(x\) is \(-\frac{4}{7} e^{-7x} + C\).

2:

To estimate the level of contamination \(N(t)\) over time \(t\), we integrate the given rate equation with respect to time:

\[ N(t) = \int \left( -\frac{3150}{t^4} - 220 \right) \, dt \]

After integration, we find:

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

To determine the constant of integration \(C\), we use the initial condition \(N(1) = 6530\) bacteria per cubic centimeter after 1 day:

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

Solving for \(C\), we find \(C = 5700\).

Therefore, the function \(N(t)\) estimating the level of contamination is:

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

3

Let \(u = 2x - 9\), then \(\frac{du}{dx} = 2\), so \(dx = \frac{1}{2} \, du\).

Substituting \(u\) and \(dx\) into the integral:

\[ \int 8.5x^u \cdot \frac{1}{2} \, du = \frac{8.5}{2} \int x^u \, du = 4.25 \int x^u \, du \]

Now, integrating \(x^u\) with respect to \(u\):

\[ = 4.25 \cdot \frac{x^{u+1}}{u+1} + C = 4.25 \cdot \frac{x^{2x-9+1}}{2x-9+1} + C = 4.25 \cdot \frac{x^{2x-8}}{2x-8} + C \]

So, the solution to the integral is:

\[ \frac{4.25x^{2x-8}}{2x-8} + C \]

This expression represents the antiderivative of the function \(8.5x^{2x-9}\).

3

f <- function(x) {
  2 * x - 9
}

lower_bound <- 4.5
upper_bound <- 8.5

result <- integrate(f, lower_bound, upper_bound)

result
## 16 with absolute error < 1.8e-13

4

f1 <- function(x) x^2 - 2*x - 2
f2 <- function(x) x + 2

lower <- -1
upper <- 4

integrand <- function(x) abs(f1(x) - f2(x))

area <- integrate(integrand, lower, upper)

area$value
## [1] 20.83333

5

D <- 110  
S <- 8.25 
H <- 3.75  

Q_star <- sqrt((2 * D * S) / H)

num_orders_per_year <- D / Q_star

cat("Economic Order Quantity (lot size):", round(Q_star, 2), "\n")
## Economic Order Quantity (lot size): 22
cat("Number of orders per year:", round(num_orders_per_year, 2), "\n")
## Number of orders per year: 5