0.1 1. Use integration by substitution to solve the integral below.

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

\(u =-7x\)

\(\frac{du}{dx} = -7\)

\(dx = -\frac{1}{7}du\)

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

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

Sub -7x for u: \(-\frac{4}{7}e^{7x}\)

0.2 2. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of dN/dt = -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.

  1. $dN = (--220)dt $ \(N(t) = \int -\frac{3150}{t^4}dt - \int220dt\)

  2. \(\int -\frac{3150}{t^4}dt = 630t^{-3}+C1\) \(\int220dt = 220t+C2\)

  3. \(N(t) = 630t^{-3}-220t + C\)

If level of contamination after 1 day N(1) was 6530, plug into equation.

\(6530 = 630(1)^{-3}-220(1) + C\) \(C = 6530-630+220 = 5900\)

Equation is then \(N(t) = 630t^{-3}-220t+5900\)

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

In the figure in the assignment, the rectangles begin at 4.5 and end at 8.5.

f <- function(x) x*2 - 9
area = integrate(f, lower = 4.5, upper = 8.5)
area
## 16 with absolute error < 1.8e-13

0.4 4. Find the area of the region bounded by the graphs of the given equations. y = x2 - 2x - 2, y = x + 2

We need to figure out where these lines intercept at the lower and upper bound on the x axis in order to integrate and solve for the area.

If you substitute y in the first equation to solve for x. \(x^2-2x-2 = x+2\) = \(x^2-3x-4 = 0\) Factors into:

\((x-4)(x+1)\) which solves x at the lower and upper bounds at -1 and +4.

f2 <- function(x) x*2 -x*3 - 4
area2 = integrate(f2, lower = -1, upper = 4)
area2
## -27.5 with absolute error < 3.1e-13

0.5 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.

create a function to figure out cost, c, based on number of orders, x and then set the derivative to 0 to find minimum cost \(c = 8.25x+3.75*\frac{110/x}{2}\) \(c = 8.25x+\frac{206.25}{x}\)

\(8.25-\frac{206.25}{x^2}=0\) \(8.25 = \frac{206.25}{x^2}\) \(x^2 = \frac{206.25}{8.25} = 25\) \(x=5\)

Num orders = 5, with 110 flat irons, gives a lot size of 22

0.6 6. Use integration by parts to solve the integral below.

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

formula is \(\int udv = uv - \int vdu\)

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

sub in values:

\(\int ln(9x)*x^6 = ln(9x)*\frac{1}{7}x^7 - \int \frac{1}{x}*\frac{1}{7}x^7\) \(ln(9x)*\frac{1}{7}x^7 - \int \frac{x^6}{7}\) \(\frac{ln(9x)x^7}{7} - \frac{x^7}{49}\) \(\frac{x^7(7ln(9x)-1)}{49}\)

0.7 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) = 1/6x

Must be non-negative, which it is because the numerator = 1 and the denominator will be positive when x is greater than 0

The total area must also be equal to 1 which would be the integral \(\int_1^{e^6} \frac{1}{6x}dx\) \(= \frac{1}{6} \int_1^{e^6} \frac{1}{x}dx\) \(=\frac{1}{6}[ln|x|]_1^{e^6}\) \(=\frac{1}{6}[ln(e^6)-ln(1)]\) \(=\frac{1}{6}[6-0] = 1\)

Therefore \(f(x) = \frac{1}{6x}\) is a PDF on interval \([1,e^6]\)