1. Use integration by substitution to solove the integration below.

\(\int4e^{-7x}dx\)

Solution: Let \(u = -7x\) by defferentiation on both sides we get this: \(\frac{du}{dx}=-7\) \(dx=\frac{-1}{7}du\)

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

  1. 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.

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 \(N'dt = \int(-\frac{3150}{t^4}-220)dt\) \(N(t)=(1050t^{-3}-220t+C)\)

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: \(N(t)=(\frac{1050}{t^2}-220t+5700)\)

  1. Find the total area of the red rectangles in the figure below, where the equation of the line is \(f ( x ) = 2x - 9\)
#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
  1. Find the area of the region bounded by the graphs of the given equations.
    \(y=x^2-2x-2, y=x+2\)
#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(area)
## [1] 20.8333
  1. 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.

Let \(C\) be cost, \(n\) be the number of orders per year and \(x\) be the number of irons in an order (lot size). \(n*x = 110\) \(x=\frac{110}{n}\) \(C=8.25*n+3.75*\frac{x}{2}\) \(C=8.25*n+3.75*\frac{\frac{110}{n}}{2}\) \(C=8.25*n+\frac{412.5}{2n}\) \(C=8.25*n+\frac{206.25}{n}\) \(C'=8.25-206.25/n^2\)

Finding the value \(n\): \(8.25-206.25/n^2=0\) \(n^2=\frac{206.25}{8.25}\) \(n=\sqrt{25}=5\) \(x = \frac{110}{5} = 22\)

Therefore lot size \(x = 22\) and the number of orders per year \(n = 5\)

  1. Use integration by parts to solve the integral below.

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

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: \(\int udv = uv-\int vdu\) \(u\frac{dv}{dx}dx = uv - \int v\frac{du}{dx}dx\) Now we can substitute the following: \(\int ln(9x). x^6 dx= ln(9x)(\frac{1}{7}x^7)-\int \frac{1}{7}x^7\frac{1}{x}dx\) \(ln(9x)(\frac{1}{7}x^7)-\int \frac{1}{7}x^7\frac{1}{x}dx\) \(\frac{ln(9x)x^7}{7}-\int \frac{x^6}{7}dx\) \(\frac{ln(9x)x^7}{7}-\frac{1}{7} \int x^6 dx\) \(\frac{ln(9x)x^7}{7}-\frac{1}{7}(\frac{1}{7}x^7+C)\) \(\frac{1}{7}x^7ln(9x)-\frac{1}{49}x^7+C\)

  1. Determine whether \(f ( x )\) is a probability density function on the interval \([1, e^6]\) . If not, determine the value of the definite integral. \(f(x) = \frac{1}{6x}\)

\(\int_{1}^{e^6} \frac{1}{6x}dx = \frac{1}{6}ln(x)|_{1}^{e^6}\) \(=\frac{1}{6}ln(e^6) - \frac{1}{6}ln(1)\) \(\frac{1}{6}*6-\frac{1}{6}*0\) \(=1\)

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.