Use integration by substitution to solve the integral below.
\(\int 4e^{-7x}dx\)
First we have to put the integral into a form like:
\(c\int e^uu'dx\)
Which for our case would be:
\(\frac{-4}{7}\int e^{-7x}(-7)dx\)
Once we substitute u and du it’s simpler to solve:
\(\frac{-4}{7}\int e^udu\)
which evaluates to:
\(\frac{-4}{7}(e^{u}+c)\)
substituting back out the u and simplifying we get:
\(\frac{-4}{7}e^{-7x}+c\)
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
To get the formula \(N(t)\) we need to integrate both sides of the contaminate rate equation. We’ll have an infinite number of solutions because of the \(+c\) term; however we’ll be solving for \(c\) knowing that \(N(1)=6530\).
By integrating we get the following formula (it helps to represent \(\frac{1}{t^4}\) as \(t^{-4}\) first)
\(N(t)=1050t^{-3}-220t+c\)
Substituting t=1 and N(t)=6530 we get:
\(6530=1050-220+c\)
So \(c=5700\)
And the level of contamination formula is:
\(N(t)=1050t^{-3}-220t+5700\)
Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x)=2x-9\).
I can’t display the image so I’ll describe it as four rectangles, each one unit across, centered on 5, 6, 7, and 8, going from the x-axis to the line of the given equation.
We can solve it by calculating the heights of the four rectangles (multiply by their widths of one) and add them together.
\(2*5-9 = 1\) \(2*6-9 = 3\) \(2*7-9 = 5\) \(2*8-9 = 7\)
Add the areas of the four rectangles together to get 16.
\(1+3+5+7=16\)
I get the same answer if I evaluate the following integral. Since it’s a straight line the triangles above the equation’s line equal the triangles below the equation’s line. But we won’t evaluate here because we already evaluate a finite integral in the next problem.
\(\int_{4.5}^{8.5}(2x-9)dx\)
Find the area of the region bounded by the graphs of the given equations.
\(y=x^2-2x-2\) \(y=x+2\)
\(x^2-2x-2=x+2\)
which equals
\(x^2-3x-4=0\)
or
\((x-4)(x+1)=0\)
So the two equations intersect at \(x=\{-1,4\}\)
We’ll evaluate both at \(x=0\) since \(0\) is between the two points of intersection.
\((0)^2-2(0)-2 < (0)+2\)
Since \(x+2\) is higher we’ll set up our integral as follows:
\(\int_{low}^{high}(higher\ equation - lower\ equation)dx\)
\(\int_{-1}^{4}((x+2)-(x^2-2x-2))dx\)
\(\int_{-1}^{4}(x+2-x^2+2x+2)dx\)
\(\int_{-1}^{4}(-x^2+3x+4)dx\)
\((-\frac{1}{3}x^3+\frac{3}{2}x^2+4x)\bigg\rvert_{-1}^4\)
\((-\frac{1}{3}(4)^3+\frac{3}{2}(4)^2+4(4))-(-\frac{1}{3}(-1)^3+\frac{3}{2}(-1)^2+4(-1))\)
\((-\frac{64}{3}+\frac{48}{2}+16)-(\frac{1}{3}+\frac{3}{2}-4)\)
\((-\frac{64}{3}+\frac{48}{2}+16)-(\frac{1}{3}+\frac{3}{2}-4)\)
\(-\frac{65}{3}+\frac{45}{2}+20\)
\(-\frac{130}{6}+\frac{135}{6}+\frac{120}{6}\)
\(\frac{125}{6}= 20.8333\)
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.
There are 8 possible combinations of lot size and order numbers that add up to 110.
CostofBeauty <- data.frame(LotSize=c(1,2,5,10,11,22,55,110),NoOfOrders=c(110,55,22,11,10,5,2,1))
CostofBeauty## LotSize NoOfOrders
## 1 1 110
## 2 2 55
## 3 5 22
## 4 10 11
## 5 11 10
## 6 22 5
## 7 55 2
## 8 110 1The ordering cost for the year is \(\$8.25 * Number\ of \ Orders\)
CostofBeauty$OrderCost <- CostofBeauty$NoOfOrders*8.25
CostofBeauty## LotSize NoOfOrders OrderCost
## 1 1 110 907.50
## 2 2 55 453.75
## 3 5 22 181.50
## 4 10 11 90.75
## 5 11 10 82.50
## 6 22 5 41.25
## 7 55 2 16.50
## 8 110 1 8.25Let’s assume we sell one unit every 3.31818 days (365/110).
Let’s assume it costs $0.01027 to store one unit for one day ($3.75/365).
Let’s assume that every order is timed so the first order is sold immediately and not stored. So if the lot size is two, every time we order, the first one sells immediately and the second we store for 3.318 days. If the lot size is five, every time we order, the first one sells immediately, the second we store for 3.3 days, the third for 6.6 days, the fourth for 9.9 days, and the fifth for 13.2 days.
Then the formula for Storage Cost is:
\(StorageCost=LotSize*(LotSize-1)/2*NoOfOrders*3.75/365\)
CostofBeauty$StorageCost <- CostofBeauty$LotSize*(CostofBeauty$LotSize-1)/2*CostofBeauty$NoOfOrders*3.75/365
CostofBeauty## LotSize NoOfOrders OrderCost StorageCost
## 1 1 110 907.50 0.0000000
## 2 2 55 453.75 0.5650685
## 3 5 22 181.50 2.2602740
## 4 10 11 90.75 5.0856164
## 5 11 10 82.50 5.6506849
## 6 22 5 41.25 11.8664384
## 7 55 2 16.50 30.5136986
## 8 110 1 8.25 61.5924658CostofBeauty$InventoryCost <- CostofBeauty$OrderCost + CostofBeauty$StorageCost
CostofBeauty## LotSize NoOfOrders OrderCost StorageCost InventoryCost
## 1 1 110 907.50 0.0000000 907.50000
## 2 2 55 453.75 0.5650685 454.31507
## 3 5 22 181.50 2.2602740 183.76027
## 4 10 11 90.75 5.0856164 95.83562
## 5 11 10 82.50 5.6506849 88.15068
## 6 22 5 41.25 11.8664384 53.11644
## 7 55 2 16.50 30.5136986 47.01370
## 8 110 1 8.25 61.5924658 69.84247The lowest inventory cost is with a lot size of 55 and two orders this year.