WebTest Name: (Test) 1. Use integration by substitution to solve the integral below.

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

\(u=-7x\) then \(du=-7dx => dx=\frac{du}{-7}\) so we have the integral \(\int 4e^{u} \frac{du}{-7}\) => \(\frac{4}{-7}e^{-7x}+C\)

  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.

\(\int\frac{-3150}{t^4}-220 dt\) = \(\frac{-3150}{-3t^3}-220t+C\) => \(6530=\frac{-3150}{-3(1)^3}-220(1)+C\) => \(C=7800\) so \(N(t)=\frac{-3150}{-3(1)^3}-220(1)+7800\)

  1. Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x) = 2x-9\). Each rectangle has a width of 1, and heights of 1, 3, 5, and 7, so the Area = 16.

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

\(x^2 - 2x - 2= x + 2\) => \(x=-1, 4\) are the points where the graphs intersect. So \(\int_{-1}^4 x + 2 - \int_{-1}^4x^2 - 2x - 2= \frac{x^2}{2}+2x - (\frac{x^3}{3}-x^2-2x) ]_{-1}^4=14.16667\)

  1. 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 C=cost, n=number of orders, and x=number of irons in the order. So nx=110, and at any given time, there are on average \(\frac{x}{2}\) irons in storage.

So \(C=\frac{x}{2}*3.75+8.25*n\) and since nx=110, then \(x=\frac{110}{n}\), so \(C=\frac{\frac{110}{n}}{2}*3.75+8.25*n\) To find the minimun point of this cost equation, we need to take the derivative and solve for 0.

\(C'=\frac{-206.25}{n^2}+8.25\) => \(0=\frac{-206.25}{n^2}+8.25\) => \(-8.25=\frac{-206.25}{n^2}\) => \(n^2=\frac{-206.25}{-8.25}\) => \(n=5\)

So the number of orders is 5, and the number of irons per order is 22.

  1. Use integration by parts to solve the integral below. \(\int ln(9x)x^6 dx\)

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

So \(\int ln(9x)x^6 dx=\int udv=uv-\int vdu=ln(9x)\frac{x^7}{7}-\int \frac{x^7}{7}\frac{1}{x}dx=\frac{x^7}{7}(ln(9x)-\frac{1}{7})+C\)

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

\(\int_1^{e^6}\frac{1}{6x}dx=\frac{1}{6}ln(x)]_1^{e^6}=\frac{1}{6}ln(e^6)-\frac{1}{6}ln(1)=\frac{6}{6}-0=1\), so f(x) is a probability density function on the interval \([1, e^6]\).