Please refer to the Assignment 13 Document.

1 Problem Set 1

Use Integration by substitution to solve the integral below.

\[\int 4e^{-7x}dx\]

1.1 Answer

Let \(F(x)=\int 4e^{-7x}dx\), \(u=-7x\).

Then \(du=-7dx\), \(dx=\frac{du}{-7}\).

Thus,

\[\int 4e^{-7x}dx = \int \frac{4}{-7}e^{u}du\] \[=-\frac{4}{7}e^{u}+C, \;where\;c\;is\;a\;constant.\] \[=-\frac{4}{7}e^{-7x}+C, \;where\;c\;is\;a\;constant.\]

2 Problem Set 2

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.

2.1 Answer

We have, \[N(t)-N(0) = \int_{0}^{t}(-\frac{3150}{s^{4}}-220)ds\]

\[= \int_{0}^{t}-3150s^{-4}-220ds\] \[=\left [ -\frac{3150s^{-3}}{-3}-220s \right ]^{t}_{0}\] \[=1050t^{-3}-220t -0+0\] \[N(t)-N(0)=1050t^{-3}-220t\]

Given \(N(1)=6530\), \(N(1)-N(0)=1050-220=830\),

\[N(0)=6530-830=5700\] Therefore, \[N(t)=1050t^{-3}-220t+5700\]

3 Problem Set 3

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

3.1 Answer

By looking at the graph given above, the area of the read rectangles \(=1+3+5+7=16\).

By integration, the area is:

\[=\int_{4.5}^{8.5}2x-9dx\] \[=\left [ x^{2}-9x \right ] ^{8.5}_{4.5}\]

\[=8.5^{2}-9 \cdot 8.5 - 4.5^{2} + 9 \cdot 4.5\]

\[=16\]

4 Problem Set 4

Find the area of the region bounded by the graphs of the given equations, \[y = x^{2}-2x-2, y = x + 2\]

4.1 Answer

Graphing the two equations:

Determining the intersections:

\[x^{2}-2x-2 = x+2\] \[x^{2}-3x-4 = 0\] \[(x+1)(x-4)=0\] \[x=-1\;or\;x=4\]

By integration,

\[\int_{-1}^{4}x+2-(x^{2}-2x-2)\] \[=\int_{-1}^{4}-x^{2}+3x+4\,dx\] \[=\left[ -\frac{x^{3}}{3}+\frac{3x^{2}}{2}+4x \right]^{4}_{-1}\] \[=-\frac{4^{3}}{3}+\frac{3\cdot4^{2}}{2}+4^{2}-(\frac{1}{3}+\frac{3}{2}-4)\] \[=-\frac{64}{3}+24+16-\frac{1}{3}-\frac{3}{2}+4\] \[\approx 20.8333\]

5 Problem Set 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.

5.1 Answer

Let \(s\) be lot size, \(n\) be number of orders, \(c\) be the inventory costs.

Assume half of inventory keep in stocks.

We have,

\[c = 8.25 \cdot n + 3.75 \cdot \frac{s}{2} \;\; and \;\; ns=110\] \[c = 8.25 \cdot n + 3.75 \cdot \frac{110}{2n}\] \[c = 8.25 \cdot n + \frac{206.25}{n}\] \[c'=8.25-206.25\cdot n^{-2}=0\] \[n=5\] \[s=\frac{110}{5}=22\] \[c=8.25 \cdot 5 +\frac{206.25}{5}=82.5\]

Therefore, to minimize the inventory costs, the lot size \(22\) and number of orders \(5\) can come up with the minimum inventory costs $82.50.

6 Problem Set 6

Use integration by parts to solve the integral below.

\[\int ln(9x)\cdot x^{6} \, dx\]

6.1 Answer

Let \(u=ln(9x)\) and \(dv=x^{6}dx\),

we have \(du=\frac{1}{9x}\cdot 9dx=\frac{dx}{x}\) and \(v=\frac{x^{7}}{7}\).

To solve the given equation using integration by parts,

\[\int ln(9x)\cdot x^{6} \, dx\]

\[=\int u \,dv = u \cdot v - \int v \,du\]

\[=ln(9x)\cdot\frac{x^{7}}{7} - \int \frac{x^{7}}{7} \frac{dx}{x}\] \[=ln(9x)\cdot\frac{x^{7}}{7} - \int \frac{x^{6}}{7}dx\] \[=ln(9x)\cdot\frac{x^{7}}{7} - \frac{x^{7}}{7\cdot7} +C\] \[=\frac{x^{7}}{7}\cdot \left[ ln(9x)-\frac{1}{7} \right] +C\]

, where C is a constant.

7 Problem Set 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. \[f(x)=\frac{1}{6x}\]

7.1 Answer

For a function to be a probability density function on the interval:

  • requirement 1: The value must always be non-negative.

  • requirement 2: The definite integral has to equal 1.

First, \(f(x)=\frac{1}{6x} >0 \;\forall \;x>0,\) so the first requirement holds on the interval \([1, e^{6}]\).

Second,

\[\int_{1}^{e^{6}} \frac{1}{6x} \,dx\] \[=\left[ \frac{ln(x)}{6} \right ] ^{e^{6}}_{1}\] \[=\frac{ln(e^{6})}{6} - \frac{ln(1)}{6}\] \[=1-0=1\]

\(\because\) The two requirements are both met.

\(\therefore\) The given function is a probability density function on the interval \([1, e^{6}]\).