Data 605 - Assignment 13
Hazal Gunduz
1. Use integration by substitution to solve the integral below.
\(\int{4e^{-7x}dx}\)
u = -7x du = -7dx \(-\frac{du}{7} = dx\)
\(\frac{-4}{7}\int{e^{u}du}\)
\(\frac{-4}{7}e^u + c\)
\(\frac{-4}{7}e^{-7x} + c\)
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.
\(\int{(\frac{-3150}{t^4} - 220}dt\)
\(\int{({-3150t^{-4}} - 220)}dt\)
\(\int{-3150t^{-4}dt - \int -220}dt\)
\(-3150 \int t^{-4}dt - 220 \int dt\)
\(-3150 \frac{-1}{3}t^{-3} - 220t\)
\(N(t) = \frac{1050}{t^{3}} - 220t + c\)
\(6530 = \frac{1050}{1^{3}} - 220(1) + c\)
6530 = 1050 - 220 + c
6530 - 1050 + 220 = c
5700 = c
\(N(t) = \frac{1050}{t^{3}} - 220t + 5700\)
3. Find the total area of the red rectangles in the figure below, where the equation of the line is f(x) = 2x - 9.
f(x) = 2x - 9
\(\int_{4.5}^{8.5} (2x - 9)dx\)
(2 * \(x^2\) - 9x) \(|_{4.5}^{8.5}\)
\((8.5^2 - 9 * (8.5)) - (4.5^2 - 9 * (4.5))\)
= 16
4. Find the area of the region bounded by the graphs of the given equations.
\(y = x^2 - 2x - 2, y = x + 2\)
Enter your answer below.
y1 <- function(x) {x^2 - (2 * x) -2}
y2 <- function(x) {x + 2}
curve(y1, from = -6, to = 6, col = 2)
curve(y2, from = -6, to = 6, add = TRUE, col = 4)
integrate(y1,-1, 4)## -3.333333 with absolute error < 1.2e-13integrate(y2,-1, 4)## 17.5 with absolute error < 1.9e-13area <- 17.5 - (-3.33)
area## [1] 20.835. 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.
n = number of orders per year
x = number of flat iron to order per year
x / 2 = average inventory level
Store cost = 3.75 * (x / 2) = 1.875x
Order cost = 8.25 * (110 / x) = 907.5 / x
Inventory cost = 1.875x + (907.5 / x)
1.875 - 907.5 * x - 2 = 0
x = 22
orders per year = 110 / n = 22
n = 5
6. Use integration by parts to solve the integral below.
\(\int In(9x) . x^{6}dx\)
uv -\(\int\) vdu
u = In(9x)
du = \(\frac{1}{x}dx\)
dv = \(x^6\)
v = \(\frac{1}{7}x^7\)
\(\frac{x^7In(9x)}{7} - \int \frac{1}{7}x^7 \frac{1}{x}dx\)
\(\frac{x^7In(9x)}{7} - \frac{1}{7} \int \frac{x^7}{x}dx\)
\(\frac{x^7In(9x)}{7} - \frac{1}{7} \int {x^6}dx\)
\(\frac{x^7In(9x)}{7} - \frac{1}{7} (\frac{x^7}{7}) + C\)
\(\frac{x^7In(9x)}{7} - \frac{x^7}{49} + C\)
7. Determine whether f(x) is a probability density function on the interval \(\left[1, e^6 \right]\). If not, determine the value of the definite integral.
\(f(x) = \frac{1}{64}\)
\(\int_{1}^{e^6} \frac{1}{6x}dx\)
\(\frac{1}{6} \int_{1}^{e^6} \frac{1}{x}dx\)
\(\frac{1}{6} (In(e^6) - In(1))\)
\(\frac{1}{6}(6 - 0) = 1\)
RPubs = > https://rpubs.com/gunduzhazal/1035657