DATA 605 HW13

Homework 13

Question #1

Solution

\(\int 4e^{-7x}dx\)
Let \(u\) = \(-7x\)
By differentiating on both sides we get this: \(\frac{du}{dx}\) = \(-7\)
\({dx}\) = \(\frac{-1}{7}du\)

Therefore, this problem can be written in a different form:
\(\int 4e^{u}.\frac{-1}{7}du\)
\(\frac{-4}{7}\int e^{u}du\)
\(\frac{-4}{7} (e^{u} + C)\)
By substituting the value of \(u\), we get the following:
\(\frac{-4}{7} e^{-7x} + C\)

Question #2

Solution

\(\frac{dN}{dt} = -\frac{3150}{t^4}-220\)
\(N^{'} = -\frac{3150}{t^4}-220\)

Integrate on both sides to get the value of \(N\)
\(\int N^{'} dt = \int (-\frac{3150}{t^4}-220) dt\)
\(N(t) = (1050t^{-3} -220t + C)\)

After day \(1\), there were \(6530\) bacteria per cubic centimeter. So, we subtitute this in the above equation for \(t = 1\): \((1050*(1)^{-3} -220*(1) + C) = 6530\)
\(C = 6530-1050+220\)
\(C = 5700\)

We substitute this value back in the equation to get the function \(N(t)\) to estimate the level of contamination:
\(N(t) = (\frac{1050}{t^3} -220t + 5700)\)

Question #3

Solution

\(Area_1 = width * length = (5.5-4.5) * (1-0) = 1*1 = 1\)
\(Area_2 = width * length = (6.5-5.5) * (3-0) = 1*3 = 3\)
\(Area_3 = width * length = (7.5-6.5) * (5-0) = 1*5 = 5\)
\(Area_4 = width * length = (8.5-7.5) * (7-0) = 1*7 = 7\)
\(Area = Area_1 + Area_2 + Area_3 + Area_4\)
\(Area = 1 + 3 + 5 + 7 = 16\)
\(Area = 16\)

\(Area = \int_{4.5}^{8.5} (2x-9) dx\)
\(Area = (2*\frac{x^2}{2}-9x) \vert_{4.5}^{8.5}\)
\(Area = (8.5^2-9*(8.5))-(4.5^2-9*(4.5))\)
\(Area = (72.25-76.5-20.25+40.5)\)
\(Area = 16\)

#Find area in-build function
f3 = function(x) {2*x -9}

#Find the difference between areas under the curve
area3 <- integrate(f3, 4.5, 8.5)$value
area3 <- round(as.numeric(area3))
print(area3)

## [1] 16

Question #4

Solution

To find the area of the region bounded by the graphs of the given equations, we first draw the graphs. The graphs representing equations intersect at \(x\) = -1 and \(x\) = 4. We determine that we need to determine the area of the shaded region.

\(Area\) = \(\int_{-1}^{4} (x+2)dx - \int_{-1}^{4} (x^2-2x-2)dx\)
\(Area\) = \(\int_{-1}^{4} [(x+2) - (x^2-2x-2)] dx\)
\(Area\) = \(\int_{-1}^{4} [(x+2-x^2+2x+2)] dx\)
\(Area\) = \(\int_{-1}^{4} [(-x^2+3x+4)] dx\)
\(Area\) = \([-\frac{x^3}{3}+\frac{3x^2}{2}+4x]\vert_{-1}^{4}\)
\(Area\) = \((-\frac{4^3}{3}+\frac{3.4^2}{2}+4.4) - (-\frac{(-1)^3}{3}+\frac{3(-1)^2}{2}+4.(-1))\)
\(Area\) = \((-\frac{64}{3}+\frac{48}{2}+16) - (\frac{1}{3}+\frac{3}{2}-4)\)
\(Area\) = \((-\frac{64}{3}-\frac{1}{3}+\frac{48}{2}-\frac{3}{2}+16+4)\)
\(Area\) = \((-\frac{65}{3}+\frac{45}{2}+20)\)
\(Area\) = \((-21.6666+22.50+20)\)
\(Area\) = \(20.8334\) square units

R Language Solution

#Find area in-build function
f1 = function(x) {x + 2}
f2 = function(x) {x^2 -2*x -2}

#Find the difference between areas under the curve
area1 <- integrate(f1, -1, 4)
area2 <- integrate(f2, -1, 4)
area <- round((area1$value - area2$value),4)
print(c(area1$value, area2$value))

## [1] 17.500000 -3.333333

print(area)

## [1] 20.8333

Question #5

Solution

Let \(C\) be cost, \(n\) be the number of orders per year and \(x\) be the number of irons in an order (lot size).

\(n*x = 110\)
\(x = \frac{110}{n}\)

\(C = 8.25*n + 3.75*\frac{x}{2}\)
\(C = 8.25*n + 3.75* \frac{\frac{110}{n}}{2}\)
\(C = 8.25*n + \frac{412.5}{2n}\)
\(C = 8.25*n + \frac{206.25}{n}\)
\(C'= 8.25 -206.25/n^2\)

Determine the value of \(n\) by equating this expression to \(0\).
\(8.25 -206.25/n^2 = 0\)
\(n^2 = \frac{206.25}{8.25}\)
\(n = \sqrt{25} = 5\)
\(x = \frac{110}{5} = 22\)
Therefore lot size \(x\) = \(22\) and the number of orders per year \(n\) = \(5\)

Question #6

Solution

\(\int ln(9x).x^6dx\)

Let \(u = ln(9x)\) and \(v = \frac{1}{7}x^7\)

\(\frac{du}{dx} = \frac{d}{dx}(ln(9x)) = \frac{1}{x}\)
\(\frac{dv}{dx} = \frac{d}{dx}(\frac{1}{7}x^7) = x^6\)

We can solve this by using integration by parts formula:
\(\int udv = uv - \int vdu\)
\(\int u\frac{dv}{dx}dx = uv - \int v\frac{du}{dx}dx\)
So, we can substitute values and get the following:
\(\int ln(9x).x^6dx\) = \(ln(9x) (\frac{1}{7}x^7) - \int \frac{1}{7}x^7\frac{1}{x}dx\)
\(ln(9x) (\frac{1}{7}x^7) - \int \frac{1}{7}x^7\frac{1}{x}dx\)
\(\frac{ln(9x) x^7}{7} - \int \frac{x^6}{7} dx\)
\(\frac{ln(9x) x^7}{7} - \frac{1}{7}\int x^6 dx\)
\(\frac{ln(9x) x^7}{7} - \frac{1}{7}(\frac{1}{7}x^7+C)\)
\(\frac{1}{7}x^7ln(9x) - \frac{1}{49}x^7 + C\)

Question #7

Solution

\(f(x) = \frac{1}{6x}\)
The integral of a PDF must be equal to \(1\):

\(\int_{1}^{e^6} f(x) dx = \int_{1}^{e^6} \frac{1}{6x} dx\)
\(\int_{1}^{e^6} f(x) dx = \int_{1}^{e^6} \frac{1}{6x} dx\)
\(\int_{1}^{e^6} f(x) dx = \frac{1}{6} \int_{1}^{e^6} \frac{1}{x} dx\)
\(\int_{1}^{e^6} f(x) dx = \frac{1}{6} ln(x) \vert_{1}^{e^6}\)
\(\int_{1}^{e^6} f(x) dx = \frac{1}{6} (ln(e^6) - ln(1))\)
\(\int_{1}^{e^6} \frac{1}{6x} dx = 1\)

In this case the sum of the definite integral for the function \(f(x) = \frac{1}{6x}\) on the interval \([0, e^6]\) is 1. Therefore, we can say that the function \(f(x)\) is a probability density function.