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

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

\(u = -7x\enspace \:\;\;\;\;\; du = -7dx\)

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

\(\int4e^u\cdot\frac{du}{-7}\) = \(-\frac{4}{7}\int e^u du\) = -\(\frac{4}{7}e^u +c\) =

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



  1. Find the function N(t) to estimate the level of contamination.

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

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

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

\(6350 = -\frac{1050}{1^3} -200(1)+ C\)

7400 = C

\(N(t) = \frac{1050}{t^3}- 220t + 7400\)



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

\(Area=f(5)+f(6)+f(7)+f(8)\)

\(Area=16\)

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

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

#Find intersection:
curve(x+2, from=-10, to=10, n=300,  
             col="blue", lwd=2,   )
curve(x^2-2*x-2, from=-10, to=10, n=300, 
             col="blue", lwd=2, add =TRUE  )

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

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

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

\(\int_{-1}^{4}(x+2)-(x^2 - 2x-2)\)

\(\int_{-1}^4(-x^2+3x+4)dx\)

20.833

  1. Find the lot size and the number of orders per uyear that will minimize inventory costs.

x = number of orders

Cost = 8.25x +3.75$

Cost = 8.25x + $

Finding the derivative and setting to zero to find critical value.

Cā€™ = 8.25 - \(\frac{206.25}{x^2}\)

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

x = 22

The number of orders per year that will minimize inventory costs is 5 so lot size is 22.

  1. Use integration of parts to solve \(\int ln(9x)x^6dx\).

\(\int uv dx\) = \(uv -\int v du\)

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

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

\(ln(9x)\frac{1}{7}x^7\) - \(\frac{1}{49}x^7\)

  1. Determine whether f(x) is a probability density function on the interval \([1,e^6]\)

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

\(\int^{e^6}_{1} \frac{1}{6x}dx\)

\(\frac{1}{6}ln(x)|^{e^6}_1\) = \(\frac{1}{6}ln(e^6) - \frac{1}{6}ln(1)\) = 1 - 0 = 1

f(x) is a probability density funciton as the area underneath the graph is equal to 1.