Use integration by substitution to solve the integral below \(\int 4e^{-7x}dx\)

I selected \(u=-7x\) and thus \(du=-7dx\). This allows for u substitution to be performed on the original equation

  1. \(4\int e^{-7x}dx\)
  2. \(-\frac{4}{7}\int e^u dx\)
  3. \(-\frac{4}{7}e^u\)

Substituting back from the original equation give the final answer

  1. \(-\frac{4}{7}e^{-7x}\)

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.

A function for the number of bacteria on a given day can be found by integrating the original function.

  1. \(\int\frac{dN}{dt}=\int-3150t^{-4}-220 dt\)
  2. \(N(t) = \frac{-3150t^{-3}}{-3}-220t+C\)
  3. \(N(t) = 1050t^{-3}-220t+C\)

Now I can use the provided piece of information that \(N(1)=6530\). Substitute into the equation to find \(C\)

  1. \(N(1) = 1050(1)^{-3}-220(1)+C\)
  2. \(C=5700\)

Finally, this value can be put back into equation from step 3 for the final answer

  1. \(N(t)=1050t^{-3}-220t+5700\)

Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x)=2x-9\).

There are four rectangles whose areas are, from left to right:

  1. \(1\times 1 = 1\)
  2. \(1\times 3 = 3\)
  3. \(1\times 5 = 5\)
  4. \(1\times 7 = 7\)

The total area is \(1+3+5+7=16\)

Find the area of the region bounded by the graphs of the given equations. \(y=x^2-2x-2\), \(y=x+2\)

funct.1 <- function(x) {x**2 -2*x - 2}
funct.2 <- function(x) {x + 2}

ggplot(data=data.frame(x=0), mapping=aes(x=x)) +
  stat_function(fun = funct.1, color='red') +
  stat_function(fun = funct.2, color='blue') +
  xlim(-5, 5)

To find the solution I will find the area under the blue line subtracting by the area under the red line for the values between the intersection points of \(x=4,-1\)

Next, I find the area underneath the upper curve of \(y=x+2\) and subtract it from the area undernearth the lower curve \(y=x^2-2x-2\) The area between the curves is \(\approx 20.833\)

integrate(funct.2, lower=-1, upper=4)
## 17.5 with absolute error < 1.9e-13
integrate(funct.1, lower=-1, upper=4)
## -3.333333 with absolute error < 1.2e-13
17.5 - -3.33333
## [1] 20.83333

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.

The information in the problem can be summarized by two equations \(C=8.25r + 3.75\frac{x}{2}\) and \(xr=110\) where \(r\) is the number of orders and \(x\) is the lot size. Combining the equations gives:

  1. \(C=8.25\frac{110}{x}+1.875x\)
  2. \(C=907.5x^{-1}+1.875x\)

The change in cost can be found by taking the derivative

  1. \(C' = -907.5x^{-2}+1.875\)

The original function is at its minimum when the derivative is at 0.

  1. \(-907.5x^{-2}+1.875=0\)
  2. \(x=22\)

Substituting back into the original equation provides \(r\)

  1. \(r=5\)

Thus to minimize cost the number of orders should be 5 consisting of 22 flat irons each.

Use integration by parts to solve the integral below \(\int ln(9x)x^6 dx\)

Integration by parts is started by identifying \(u\) and \(\frac{dv}{dx}\) and then calculating the missing pieces

  1. \(u=ln(9x)\)

  2. \(\frac{du}{dx}=\frac{1}{x}dx\)

  3. \(\frac{dv}{dx}=x^6dx\)

  4. \(v = \frac{x^7}{7}\)

These pieces are substituted into the integration by parts formula

  1. \(uv-\int v du\)
  2. \(\frac{ln(9x)x^7}{7}-\int\frac{x^7}{7}\frac{1}{x}dx\)

The integral can be calculated more easily now

  1. \(\frac{ln(9x)x^7}{7}-\frac{1}{7}\frac{x^7}{7}\)

Finally, the result is simlfied

  1. \(\frac{x^7(7ln(9x)-1)}{49}\)

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

In order to a be a probability density function the integral of the function on the provided interval must be 1.

  1. \(\int_{1}^{e^6}\frac{1}{6x}dx\)
  2. \(\frac{1}{6}\int_{1}^{e^6}x^{-1}dx\)
  3. \(\frac{1}{6}ln(x)\bigg\rvert_{1}^{e^6}\)
  4. \(\frac{1}{6}(ln(e^6)-ln(1))\)
  5. \(\frac{1}{6}(6)\)
  6. \(1\)

The integral value of 1 has been verify and thus the function is a probability density function.