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

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

Using substitution

\(u = -7x\)

\(du/dx = d/dx (-7x) = -7\)

\(du = -7 dx\)

Hence

\(\int 4 e^{-7x} dx = \int -1/7 (-7) 4 e^{-7x} dx\)

\(= -1/7 \int 4 e^{-7x} (-7dx) = -1/7 \int 4 e^u du\)

\(= -1/7 \int 4 e^{-7x} (-7dx) = -1/7 \int 4 e^u du\)

\(= -4/7 \int e^u du = -(4/7) e^u + C\)

Replacing u with the value -7x

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

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.

Contamination N of the pond will be derived by taking the integral of the change rate: dN/dt.

\(N = \int dN/dt = \int (-(3150/t^4) - 220)\) dt

\(N = -3150 t^{-3}/(-3) - 220t + C = 1050 t^{-3} - 220t + C\)

At t = 1 day, N = 6530

Hence substituting t = 1,

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

\(C = 5700\)

Hence the function to describe the level of contamination, N at a given time t is:

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

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

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

\(Area = ({8.5}^2 - 9(8.5) + C) - ({4.5}^2 - 9(4.5) + C) = 16\)

4. Find the area of the region bounded by the graphs of the given equations.

\(y = x^2 - 2x - 2, y = x + 2\)

Plotting the 2 functions first

fun1 <- function(x)
{
  x^2 - 2*x - 2
}

fun2 <- function(x)
{
  x + 2
}

curve(fun1, -6, 6)
plot(fun2, -6, 6, add=TRUE)

Calculating the solutions, where the 2 graphs meet.

\(x^2 - 2x - 2 = x + 2\)

\(x^2 - 3x - 4 = 0\)

\((x-4)(x+1) = 0\)

Hence, the solutions are present at the x-values x = 4, and x = -1

Getting y for these values, the 2 points are (4,6) and (-1,1)

Now to get the area between the 2 curves, area will be the difference between the areas of the 2 curves, which means the difference in the integrals between x=-1 and x=4

$Area = _{-1}^4 (x+2) - (x^2 - 2x - 2) $ = 20.83

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.

Let x = no. of orders per year

y = lot size

No. of irons produced per year = x * y, which should be greater than or equal to 110 x*y >= 110

No. of extra irons, not sold, which will be added to the cost of storage, which is $3.75 per iron.

Inventory cost = 3.75 (xy - 110) + 8.75 x

For minimum inventory cost, we will minimize x, that means x = 1. Hence y = 110

Hence, for minimum inventory costs, 1 order of size 110 will be optimum.

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

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

As per the integration by parts formula:

\(\int f(x)g^{'}(x) = f(x)g(x) - \int f^{'}(x)g(x)\)

Now we will take f(x) = ln(9x), hence

\(\ f^{'}(x) = 1/x\)

\(and we take \ g^{'}(x) = x^6, hence g(x) = x^7 / 7\)

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

\(\ (x^7)ln(9x)/7\) - \(\int 1/x * (x^7)/7\) dx

\(\ = (x^7)ln(9x)/7\) - 1/7 \(\int x^6\) dx

\(\ = (x^7)ln(9x)/7\) - \(x^7 / 49 + C\)

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.

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

\(\int_1^{e^6} 1/{(6x)}\) dx =

\((1/6) (ln(e^6) - ln(1))\) =

$(1/6) (6 x 1 - 0) $ =

= 1

Hence this is a probability distribution with 100% probability under the same.