Problem Set 1

Use integration by substitution to solve the integral below.

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

Answer

Substitute \(\quad u=-7x\quad \rightarrow \quad \frac { du }{ dx } =-7\quad \rightarrow\quad dx=-7du\)

Then,

\[-\frac { 4 }{ 7 } \int { { e }^{ u } } du\]

Now solving for \(\int { { e }^{ u } } du\)

Apply the exponential rule:

\(\int { { e }^{ u } } du = { e }^{ u }\)

Plug in the solved integral:

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

Undo substitution u=-7x:

\(-\frac { 4{ e }^{ u } }{ 7 }=-\frac { 4{ e }^{ -7x } }{ 7 }\)

So, \(\int { 4{ e }^{ -7x }dx}= -\frac { 4{ e }^{ -7x } }{ 7 } + C\) where C is th constant of integration.

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.

Answer

Given the contamination is changing at a rate of \(\frac { dN }{ dt } =-\frac { 3150 }{ { t }^{ 4 } } -220\)

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

Integrate both sides:

\(\int { dN } =\int { (-\frac { 3150 }{ { t }^{ 4 } } -220) } dt\)
\(N(t)=-3150\frac { { t }^{ -3 } }{ -3 } -220t+C\)
\(N(t)=-1050({ t }^{ -3 })-220(t)+C\)

Given after 1 day, N=6530 bacteria per cubic cm. So,

\(6530=-1050({ 1 }^{ -3 })-220(1)+C\)
\(6530=-1050-220+C\)
\(C=6530+1050+220\)
\(C=7800\)

So, the function N(t) to estimate the level of contamination is

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

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.

Answer

\(Total Area=\sum _{ i=1 }^{ n }{ f({ x }_{ i })\Delta x }\)

Given f(x) = 2x - 9,

f(5) = 2(5)-9 = 1
f(6) = 2(6)-9 = 3
f(7) = 2(7)-9 = 5
f(8) = 2(8)-9 = 7

and \(\Delta x=\frac { 8-5 }{ 3 } =1\)

So, the total area of the red rectangles is (1+3+5+7)(1) = 16

Problem Set 4

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

\[y={ x }^{ 2 }-2x-2,\quad y=x+2\]

Answer

\(Area=\int { [({ x }^{ 2 } } -2x-2)-(x+2)]dx\)
\(=\int { ({ x }^{ 2 } } -3x-4)dx\)
\(=\frac { { x }^{ 2+1 } }{ 2+1 } -3\frac { { x }^{ 1+1 } }{ 1+1 } -4x+C\)
\(=\frac { { x }^{ 3 } }{ 3 } -\frac { { 3x }^{ 2 } }{ 2 } -4x+C\)

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.

Answer

Let x be the number of flat irons to order per year and x/2 is the average inventory level.

Storage Cost Per Year = 3.75 * x/2 = 1.875x
Ordering Cost Per Year = 8.25 * 110/x = 907.5/x
Inventory Cost Per Year = 1.875x + 907.5/x

So, f(x) = 1.875x + 907.5/x

To find the minimized inventory costs, we need to differentiate f(x) and solve for f’(x)=0.

f’(x) = 1.875 - 907.5/x^2

1.875 - 907.5/x^2 = 0

x = sqrt(907.5/1.875) = sqrt(484) = 22

So, the lot size for each order should be 22 flat irons and the number of orders per year is 110/22=5 orders.

Problem Set 6

Use integration by parts to solve the integral below.

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

Answer

\(\int { { x }^{ 6 }\bullet ln(9x)} dx\)

Integrate by parts:

\(\int { fg' } =fg-\int { f'g }\)

\(f=ln(9x)\rightarrow f=ln(x)+ln(9)\rightarrow f'=\frac { 1 }{ x }\)
\(g'={ x }^{ 6 }\rightarrow g=\frac { { x }^{ 7 } }{ 7 }\)

So,

\(\int { fg' } =\frac { { x }^{ 7 }ln(9x) }{ 7 } -\int { \frac { { x }^{ 6 } }{ 7 } } dx\)
\(=\frac { { x }^{ 7 }ln(9x) }{ 7 } -\frac { { x }^{ 7 } }{ 49 }+C\)
\(=\frac { { x }^{ 7 }(7ln(9x)-1) }{ 49 } +C\)

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.

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

Answer

For probability density function,

\(\int _{ -\infty }^{ +\infty }{ f(x)dx=1 }\)

Let’s now find out whether the f(x) is a PDF,

\(\int _{ 1 }^{ { e }^{ 6 } }{ \frac { 1 }{ 6x } dx=\frac { 1 }{ 6 } } \int _{ 1 }^{ { e }^{ 6 } }{ \frac { 1 }{ x } dx }\)
\(=\left| \frac { ln(x) }{ 6 } \right| \begin{matrix} { e }^{ 6 } \\ 1 \end{matrix}\)
\(=\frac { 1 }{ 6 } ln({ e }^{ 6 })-\frac { 1 }{ 6 } ln(1)\)
\(=\frac { 1 }{ 6 } (6)-\frac { 1 }{ 6 } (0)\)
\(=\frac { 6 }{ 6 }\)
\(=1\)

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