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

\(\int 4e^{-7x} \, dx\) \[ u = -7x \]

differentiate \(u\) with respect to \(x\) to find \(du\): \[ du = -7 \, dx \] \[ dx = -\frac{1}{7} \, du \]

Substituting: \[ \int 4e^{-7x} \, dx = \int 4e^u \left(-\frac{1}{7}\right) \, du \] \[ = -\frac{4}{7} \int e^u \, du \]

The integral of \(e^u\) : \[ -\frac{4}{7} \int e^u \, du = -\frac{4}{7} e^u + C \]

substitute into the original variable \(x\): \[ -\frac{4}{7} e^u + C = -\frac{4}{7} e^{-7x} + C \]

\[ \int 4e^{-7x} \, dx = -\frac{4}{7} e^{-7x} + C \]

  1. 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.

\[ \frac{dN}{dt} = -\frac{3150}{t^4} - 220 \]

\[ \int -\frac{3150}{t^4} \, dt = -3150 \int t^{-4} \, dt = -3150 \left(-\frac{1}{3} t^{-3}\right) = \frac{3150}{3t^3} \]

\[ \int -220 \, dt = -220t \]

\[ N(t) = \frac{3150}{3t^3} - 220t + C \]

Day 1 :\(N(1) = 6530\): \[ N(1) = \frac{3150}{3(1)^3} - 220(1) + C = 1050 - 220 + C \] \[ 6530 = 830 + C \] \[ C = 6530 - 830 = 5700 \]

\[ N(t) = \frac{3150}{3t^3} - 220t + 5700 = \frac{1050}{t^3} - 220t + 5700 \]

  1. Find the total area of the red rectangles in the figure below, where the equation of the line is f ( x ) = 2x-9.
f <- function(x) {
  return(2*x - 9)
}

x <- seq(4.5, 8.5, by =.5)

heights <- sapply(x, f)

width <- .5

area <- sum(heights * width)

area
## [1] 18

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

    \[ x^2 - 2x - 2 = x + 2 \]

    \[ x^2 - 2x - 2 - x - 2 = 0 \implies x^2 - 3x - 4 = 0 \] \[ (x - 4)(x + 1) = 0 \] \(x = 4\) and \(x = -1\).

    \[ A = \int_{-1}^4 [(x + 2) - (x^2 - 2x - 2)] \, dx \]\[ A = \int_{-1}^4 [x + 2 - x^2 + 2x + 2] \, dx = \int_{-1}^4 [-x^2 + 3x + 4] \, dx \]

    \[ A = \int_{-1}^4 [-x^2 + 3x + 4] \, dx = 20.83 \]

  2. 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.

Given:

Annual demand, \(D = 110\)

flat irons - Holding cost per unit per year, h = $3.75

Fixed cost per order, ( K = $8.25 )

The formula for the EOQ, \(Q\), is: \[ Q = \sqrt{\frac{2DK}{h}} \]

\[ Q = \sqrt{\frac{2(110)8.25}{3.75}} = \sqrt{\frac{1815}{3.75}}=\sqrt{484} = 22 \]

\[ 22x = 110 => x = \frac{110}{22} = 5 \]

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

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

\(u = \ln(9x)\) \(dv = x^6 \, dx\)

\(du = \frac{d}{dx}[\ln(9x)] \, dx = \frac{1}{x} \, dx\) - \(v = \int x^6 \, dx = \frac{x^7}{7}\)

\[ \int \ln(9x) \cdot x^6 \, dx = \left( \ln(9x) \cdot \frac{x^7}{7} \right) - \int \frac{x^7}{7} \cdot \frac{1}{x} \, dx \] \[ = \frac{x^7}{7} \ln(9x) - \int \frac{x^6}{7} \, dx \]\[ \int \frac{x^6}{7} \, dx = \frac{1}{7} \int x^6 \, dx = \frac{1}{7} \cdot \frac{x^7}{7} = \frac{x^7}{49} \]

\[ \int \ln(9x) \cdot x^6 \, dx = \frac{x^7}{7} \ln(9x) - \frac{x^7}{49} + C \]

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

\[ \int_1^{e^6} \frac{1}{6x} \, dx \]

\[ \frac{1}{6} \int_1^{e^6} \frac{1}{x} \, dx = \ln(x) \]

\[ \ln(e^6) - \ln(1) = 6 - 0 = 6 \]

\[ \frac{1}{6} \cdot 6 = 1 \]

\(f(x) = \frac{1}{6x}\) is probability density function on the interval \([1, e^6]\)