1

Use integration by substitution to solve the integral below.

\(\int 4e^{-7x} dx\)

Recall \(\int e^u du = e^u\).

Make \(u = -7x\) and \(du = -7 x\).

\(\int 4e^u \frac{du}{-7}\)

\(-\frac{4}{7} \int e^u du\)

\(-\frac{4}{7}e^{-7x} + C\)

2

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.

\(\frac{dN}{dt} =-\frac{3150}{t^4} - 220\)

\({dN} =(-\frac{3150}{t^4} - 220)dt\)

Let us split and integrate.

\({dN} =(-\frac{3150}{t^4} - 220)dt\)

\(N = \int-\frac{3150}{t^4}dt - \int220 dt\)

\(N(t) = \frac{1050}{t^3} - 220t + C\)

Solve for \(N(t) = 6530\) at for day 1.

\(6530 = \frac{1050}{1^3} - 220(1) + C\)

\(C = 5700\)

So:

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

3

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

Area = \(\int_{4.5}^{8.5} 2x - 9dx\)

Solve in R.

triangle <- function(x)
{
  2*x - 9
}

integrate(triangle, lower = 4.5, upper = 8.5)$value
## [1] 16

4

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

Enter your answer below.

Solving for x, \(x+ 2 = x^2 - 2x - 2\) \(x^2 - 3x - 4 = -\) \((x - 4)(x + 1)\)

\(x = 4\) or \(x = -1\)

Solve in R for area of intersection.

intersect <- function(x)
{
  return(x^2 -3*x -4)
}

integrate(intersect,lower=-1,upper=4)$value
## [1] -20.83333

Flip the sign.

5

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.

Minimizing inventory costs would be minimzing combo of storage and ordering costs. Following convention, we’ll assume half of the lot size remains in stock from the previous order. \(x\) is our lot size.

\(f(x) = 3.75\frac{x}{2} + 8.25\frac{110}{x}\)

To find minimum, solve where derivative is 0, verifying number of minimums.

\(f(x) = 1.875x - \frac{907.5}{x^2} = 0\)

\(x^2 = 907.5/1.875\)

sqrt(907.5/1.875)
## [1] 22
110/22
## [1] 5

That’s 5 orders of 22 irons.

6

Use integration by parts to solve the integral below.

\(\int ln(9x) · x^6 dx\)

Recall:

\(\int u*dv = uv - \int v*du\)

Let \(u = ln(9x)\) so \(du = 1/x\). And \(dv = X^6\) so \(v = \frac{x^7}{7}\).

\(\int ln(9x) * x^6 dx = \frac{x^7}{7} * ln(9x) - \int \frac{x^7}{7} * \frac{dx}{x}\)

\(ln(9x) * \frac{x^7}{7} - \frac{x^7}{49} + C\)

7

Determine whether f ( x ) is a probability density function on the interval 1, e 6 . If not, determine the value of the definite integral.

\(f(x) = \frac{1}{6x}\)

\(F(x) = \int_{1}^{e^6}\frac{dx}{6x}\)

\(F(x) = \frac{1}{6} \int_{1}^{e^6}\frac{dx}{6x}\)

\(F(x) = \frac{1}{6}(ln(e^6) - ln(1))\)

\(F(x) = \frac{1}{6}(6-0) = 1\)

It tis a probability density function on the stated interval.