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

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

Move constant

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

Use u-substitution

\(u = -7x\)

Solve for dx

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

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

Replace dx

\(4 \int -\frac{1}{7} e^{u}du\)

Substitute u back in and add constant

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

2) Biologists are treating a pond contaminated with bacteria…

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

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

\(N' = 3150 \frac{t^{-3}}{-3} -220t + c\)

Set equal to 6530 and substitute 1

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

\(6530 = -1270 + c\)

\(7800 = c\)

Therefore, the function is:

\(\frac{-1050}{t^3} -220t + 7800\)

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

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

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

Set equations equal to each other and solve for x:

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

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

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

Therefore, we integrate -1 to 4, the difference of the equations

func <- function(x) {x+2 - (x^2-2*x-2)}
integrate(func, lower = -1, upper = 4)
## 20.83333 with absolute error < 2.3e-13

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.

Holding cost

\(H(x) = (holding-cost-per-unit) * \frac{x}{2}\)

\(3.75 * (x/2)\)

\(1.875x\)

Reorder cost

\(R(x) = (cost-per-order) * \frac{total-units}{x}\)

\(8.25 * (110/x)\)

\(\frac{907.5}{x}\)

Total inventory cost

\(C(x) = H(x) + R(x)\)

\(1.875x + \frac{907.5}{x}\)

Find first derivative of total inventory cost function:

f=expression(1.875*x + (907.5/x))
D(f,'x')
## 1.875 - 907.5/x^2

Set equal to zero and solve for x:

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

\(x^2 = 484\)

\(x\pm22\)

Therefore, since -22 is not in our range, the lot size that will minimize costs is:

Lot size: 22

Which means the number of orders that minimizes costs is:

Orders: 5 (110/22)

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

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

\(u = ln(9x)\)

f=expression(log(9*x))
derv_1 <- D(f,'x')
Simplify(derv_1)
## 1/x

\(u' = \frac{1}{x}\)

\(v' = x^6\)

Integrating we get… \(v = \frac{x^7}{7}\)

Using the integration by parts formula: \(\int udv = uv - \int vdu\)

We get… \(ln(9x) \frac{x^7}{7} - \int \frac{1}{x} * \frac{x^7}{7} dx\)

yac_str("Integrate(x) (x^6)/7")
## [1] "x^7/49"

Simplified:

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

7) Determine whether f ( x ) is a probability density function on the interval…

Let’s check if function:

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

Integrates to one over the range [1, \(e^6\)]

func <- function(x) {1/(6*x)}
integrate(func, lower = 1, upper = exp(6))
## 1 with absolute error < 9.3e-05

It is a probability density function on the interval!

References

Minimizing Inventory Costs