Data 605 Assignment 13
Question One
Use Integration to solve the Integral below
\[∫4e−7xdx\]
We will let
\[u = -7x\quad therefore \quad du/dx = -7 \quad and \quad dx = du/-7\]
which will give us
\[-4/7\int e^udu\]
which solves to
\[-4/7e^{-7x}+C\]
Question Two
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.
\[\begin{align} \int -\frac{3150}{t^4}-220 \space dt &= -3150 \int t^{-4} \space dt - \int 220 \space dt \\ &= -3150 (\frac{-1}{3} t^{-3}) - (220t) + C \\ &= 1050 t^{-3} - 220t + C \\ \\ N(t) &= 1050 t^{-3} - 220t + C \\ N(1) &= 6530 \\ 1050 \cdot 1^{-3} - 220 \cdot 1 + C &= 6530 \\ C &= 6530 - 1050 + 220 \\ C &= 5700 \\ N(t) &= 1050 t^{-3} - 220t + 5700 \end{align}\]
Question Three
Find the total area of the red rectangles in the figure below, where the equation of the line is f(x)=2x−9.
The way I approach this problem is looking at it as 4 rectangles, that are 1X1, 1X3, 1X5, 1X 7 so therefore: \[\text{Area} = 1(1)+3(1)+5(1)+7(1) = 16\]
Question Four
Find the area of the region bounded by the graphs of the given equations. \[y = x^2-2x-2\\y=x+2\]
fun1 <- function(x){
x^2-2*x-2
}
fun2 <- function(x){
x+2
}
plot(fun1, -2, 5, col="red", ylab="y")
plot(fun2, -2, 5, col="blue", add=TRUE)
legend("top", c("x^2-2x-2", "x+2"), fill=c("red", "blue"))
First we need to find the intersection of this two equations
\[\begin{align} x^2-2x-2 &= x+2 \\ x^2-2x-x-2-2 &= 0 \\ x^2-3x-4 &= 0 \\ (x-4)(x+1) = 0 \\ x &= 4, -1 \end{align}\]
these numbers make sense from what we see above so we then set an intergral for this intersection region
\[\begin{align} \int_{-1}^4 ((x+2)-(x^2-2x-2)) \space dx &= \int_{-1}^4 -x^2+3x+4 \space dx \\ &= \left.\left( -\frac{1}{3}x^3 + \frac{3}{2}x^2 + 4x \right) \right|_{-1}^3 \\ &= \left( -\frac{1}{3}3^3 + \frac{3}{2}3^2 + 4 \cdot 3 \right) - \left( -\frac{1}{3}(-1)^3 + \frac{3}{2}(-1)^2 + 4(-1) \right) \\ &= \left( -9 + 13\frac{1}{2} + 12 \right) - \left( \frac{1}{3} + \frac{3}{2} - 4 \right) \\ &= 20 \end{align}\]
Question Five
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.
Number of orders per year = n
lot size = s
Cost = c
Assume half of inventory keep in stocks:
\[c = 8.25n + \frac{375}{2n}\] \[c = 8.25n + \frac{206.25}{n}\] \[c' = 8.25 - \frac{206.25}{n^{_{2}}}\] \[c′=0\] \[0 = 8.25 - \frac{206.25}{n^{_{2}}}\] \[n = 5\]
Number of order per year is 5
Lot size is 22 and inventory cost is 78.75.
- Use integration b
Question Six
Use integration by parts to solve the integral below
\[\int ln(9x)*x^6dx\]
\[u=ln(9x)\] \[\frac{dv}{dx}=x^{6}\] \[du = \frac{9}{9x}dx = \frac{1}{x}dx\] \[dv = x^6dx\] \[v = \frac{1}{7}x^7\] \[\int u dv= uv-\int v du\] \[= ln(9x)\frac{1}{7}x^7 - \int\frac{1}{7}x^7\frac{1}{x}dx\] \[=ln(9x)\frac{x^7}{7} - \frac{x^7}{49}-C\]
Question Seven
Determine whether f(x) is a probability density function on the interval [1, e^6]. If not, determine the value of the definite integral.
\[f(x)= \frac{1}{6x}\] \[\begin{align} \int_1^{e^6} \frac{1}{6x} dx &= \frac{1}{6} \int_1^{e^6} \frac{1}{x} dx \\ &= \left.\frac{1}{6} \ln x \right|_1^{e^6} \\ &= \frac{1}{6} (\ln e^6 - \ln 1) \\ &= \frac{1}{6} (6- 0)\\ &= 1 \end{align}\]
Since the integral results in an answer of one, f(x) is a probability density function