Use integration by substitution to solve the integral below.
\[ \int{4e^{-7x}dx}\]
\[ u=-7x,\text{ }u'=-7x\ dx = -\frac{1}{7}\\ \int{4e^{-7x}}dx = 4 \int{e^u} du * u'\\ -\frac{4}{7}\int{e^u du}\\ -\frac{4}{7} e^u\\ -\frac{4}{7} e^{-7X} \]
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.
Constant of integration, N0 if N(1) = 6530
\[\frac{dN}{dt}=-\frac{3150}{t^4} - 220\]
\[dN =(\frac{3150}{t^4}−220)dt\]
\[N=\int{\frac{3150}{t^4}dt} −\int{220dt}\]
\[N = N_0 − \frac{3150}{3t^3}− 220t\]
\[N(_1) = N_0 −\frac{1050}{1^3}−220(1)\]
\[N_0 = 6530 + 1050 + 220\]
\[N_0 = 7800\]
\[N = 7800−\frac{1050}{t^3}−220t\]
Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x) = 2x-9\).
Area is a triangle from 4.5 to 8.5.
funcArea <- function(x){
2*x - 9
}
integrate(funcArea, 4.5, 8.5)## 16 with absolute error < 1.8e-13Find the area of the region bounded by the graphs of the given equations.
\[ \begin{split} y &= x^2-2x-2 \\ y &= x+2 \end{split} \]
X = function(x) {x + 2}
Y = function(x) {x^2 -2*x -2}ax <- integrate(X, -1, 4)
yx <- integrate(Y, -1, 4)
(Area <- round((ax$value - yx$value),4))## [1] 20.8333A 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.
\[Storage\ cost/ Yr = 3.75 \times x/2 = 1.875x\]
\[Ordering\ cost/Yr = 8.25 \times 110/x = 907.5/x\]
\[Inventory\ cost = 1.875x+907.5/x = f(x)\]
\[ \begin{split} f'(x) &= 1.875-\frac{907.5}{x^2} \\ f'(x) &= 0 \end{split} \]
\[ \begin{split} 1.875-\frac{907.5}{x^2} &= 0 \\ 1.875&= \frac{907.5}{x^2} \\ 1.875x^2&= 907.5 \\ x^2&= \frac{907.5}{1.875} \\ x&= \sqrt{\frac{907.5}{1.875}} \\ x&=\sqrt{484} \\ x&=22 \end{split} \] \[ orders = 110/22=5\]
Use integration by parts to solve the integral below.
\[ \int { ln(9x).{ x }^{ 6 }dx } \]
\[=\frac{1}{7}x^7 * ln(9x) − \int{\frac{1}{7}x^7 * \frac{1}{x}dx}\]
\[=\frac{1}{7}x^7 * ln(9x) - \int{\frac{1}{7}x^6dx}\]
\[= \frac{7}{49}x^7 * in(9x) - \frac{1}{49} x^7dx + C\]
\[= \frac{1}{49}x^7*(7ln(9x) - 1) + C\]
Determine whether f (x) is a probability density function on the interval 1, e6. If not, determine the value of the definite integral.
\[f(x) = \frac{1}{6x}\]
Is the area under the curve within the interval equal to 1
\[F(x) = \int_1^{e^6}{\frac{1}{6x}dx}=1\]
\[ \begin{split} \int_1^{e^6}{\frac{1}{6x}dx} &= \frac{1}{6}ln(x) |_1^{e^6} \\ &= \frac{1}{6}ln(e^6) - \frac{1}{6}ln(1) \\ &=\frac{1}{6}\times6 - \frac{1}{6}\times 0 \\ &= 1 - 0 \\ &= 1 \end{split} \]
f(x) is a probability density function