605 Week 13

Problem 1

\[ \int 4e^{-7x}dx \implies 4\int e^{-7x}dx \] Let \[ u = -7x \implies du = -7dx \implies \\ \frac{-4}{7} \int e^u du \\ = -\frac{4e^u}{7} + C \] where $ u = -7x$, such that \[ -\frac{4}{7}e^{-7x} + C \]

Problem 2

\[ \frac{dN}{dt} = -\frac{3150}{t^4}-220 \text{ bacteria/cc/day, where t is the number of days of treatment.} \] Additionally, we know that when \(t = 1, N = 6530\). To find \(N(t)\), we must integrate our differential function such that \[ \int \frac{dN}{dt} = \\ \text{and} \\ N(t = 1) = 6530 \] \[ = \frac{1050}{t^3}-220t + C \] If we let \(t = 1\), \[ = 1050-220 + C = 6530 \implies \\ C = 5700 \] such that \[ f(N) = \frac{1050}{t^3} - 220t + 5700 \] ## Problem 3 If \[ f(x) = 2x-9 \] and the region is bounded by x = 4 and x = 8.5, we can find the area with a simple integral.

integrand <- function(x){2*x-9}
integrate(integrand, 4, 8.5)
## 15.75 with absolute error < 1.8e-13

Problem 4

We can look at these two functions in R

function1 <- function(x){x^2-2*x-2}
function2 <- function(x){x + 2}
plot(function1, -1, 4)

plot(function2, -1 , 4)

First, we must find the bounds of our integral, corresponding to the places where these functions meet. \[ x^2-2x - 2 = x + 2 \implies \\ x^2-3x-4 = 0 \implies \\ x = 4, x = -1 \] This gives us an integral such that \[ \int_{-1}^{4} ((x+2)-(x^2-2x-2))dx \\ = \int_{-1}^{4} (-x^2+4x+4) \\ = \frac{t^3}{3} + 2t^2 + 4 \Big|_{-1}^{4} \\ = \frac{85}{3} \] ## Problem 5 Given the below parameters, find the lot size and the number of orders per year that will minimize inventory costs.

n.max <- 110
storage <- 3.75
fixed <- 8.25
g <- function(x){110/x}
f <- function(n){storage*(n) + fixed *g(n)}
plot(f,from = 0, to=100)

By inspection, we can see that the optimal point is at the elbow, near 15. However, if we define our cost function as \[ C(n,x) = 8.25 n + 3.75 x \] where n is the number of lots and x is the number of widgets in each lot. We also know that \[ 110 = n \cdot x \implies \frac{110}{x} = n \] By substituting equation 2 into equation 1, we can obtain a new cost function \[ C(x) = 8.25x + \frac{3.75 * 110}{x} \implies \\ C'(x) = 3.75 - \frac{907.5}{x^2} \implies \\ x = 15.6 \] However, since we must make a discrete decision, we will check both x = 15 and x = 16. \[ C(x = 15) = 64 C(x = 16) = 60 \] The cost is minimized when there are 16 widgets per lot. ## Problem 6 Use integration by parts to solve \[ \int \ln(9x)\cdot x^6 dx \implies \\ f = \ln{9x}\\ df = \frac{1}{x}dx \\ dg = x^6 dx \\ g = \frac{x^7}{7} \implies \\ \frac{1}{7} x^7 \ln{9x} - \frac{1}{7} \int x^6 dx \implies \\ \frac{1}{49} x^7 (7\ln(9x)-1) + C \] ## Problem 7 Determine whether \(f(x)\) is a pdf on the interval [1,\(e^6\)]. If not, determine the value of the definite integral. \[ f(x) = \frac {1}{6x} \implies \\ \int_1^{e^6} \frac{1}{6x}dx \\ = \frac{1}{6} \int_1^{e^6}\frac{1}{x}dx \\ = \frac{\log{x}} \Big|_{1}^{e^6} \\ = 1 \] It satifies the conditions of a pdf on this interval.