1. Use integration by substitution to solve the integral below. \(\color{red} {\int 4e^{-7x}dx}\)

Let \(u = -7x\) or

\(\cfrac { du }{ dx } = -7\Rrightarrow\) \({ dx } = \cfrac { -1 du}{ 7 }\)

Hence, \({\int 4e^{-7x}dx}\) \(\Rrightarrow\cfrac {-4}{7}{\int e^{u}du}\)

Now solving: \({\int e^{u}du}\) \(\Rrightarrow e^u\) According to exponential rule.

Substituting this back to equation above \({\int 4e^{-7x}dx}\) \(\Rrightarrow \cfrac {-4}{7}{\int e^{u}du}\) \(\Rrightarrow \cfrac {-4}{7}e^u\)

Substituting back the value of u:- \(\cfrac {-4}{7}e^u\) \(\Rrightarrow \cfrac {-4}{7}e^{-7x}\) \(\Rrightarrow (-\cfrac {4}{7}e^{-7x} + C)\)

2. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \(\color{red} {\cfrac { dN }{ dt } = -\cfrac { 3150 }{ t^4 } - 220}\) 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.

\({\cfrac { dN }{ dt } = -\cfrac { 3150 }{ t^4 } - 220}\Rrightarrow\) \(N^{'}(t) = \int {-\cfrac { 3150 }{ t^4 } - 220}. dt \Rrightarrow\) \(\cfrac {1050} {t^3} -220t + C\)

Since, \(N(1) = 6530\) The Above Equation can be solved for value of C

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

\(C = 6530-1050+220\)

\(C = 5700\)

We can not replace the value of C to calculate \(N^{'}(t)\) \(=\cfrac {1050} {t^3} -220t +5700\)

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

## 16 with absolute error < 1.8e-13

4. Find the area of the region bounded by the graphs of the given equations. \(\color{red} {y = x^2 - 2x - 2, y = x + 2}\) Enter your answer below.

## [1] -1  4
## 20.83333 with absolute error < 2.3e-13

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.

Total number of Irons to sell = 110. Let x be a number of flat irons to order.

\(Total Yearly storage cost=Storage cost×Average number of irons stored =\) \(3.75× \cfrac{x}{2}=1.875x\)

\(Total Yearly ordering cost=Cost of each order×Number of orders =\) \(8.25×\cfrac{110}{x}=\cfrac{907.5}{x}\)

\(Total Inventory cost=Yearly storage cost+Yearly ordering cost =\)\(1.875x+\cfrac{907.5}{x}=f(x)\)

To find the minimal value, we calculate derivative and solve at 0 the above equation becomes:

\(f^{'}(x) = 1.875-\cfrac{907.5}{x^2} = 0 \Rrightarrow\) \(1.875 = \cfrac{907.5}{x^2} \Rrightarrow x = \sqrt {\cfrac{907.5}{1.875}}\)

Solving above gives \(x = 22\) Lots

Orders is calculated by:- number of irons to \(\cfrac{Iron Sales}{Lots}\) \(\Rrightarrow \cfrac{110}{22} = 5\) Orders

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

Integrate by parts formula tells us: \(\int fg^{'} = fg - \int f^{'}g\)

Let \(f = ln(9x)\) and \(g^{'} = x^6\) or \(f^{'} = \cfrac {1}{x}\) and \(g = \cfrac {x^7}{7}\)

Substituting into above equation \(\cfrac {x^7 ln(9x)}{7} - \int \cfrac{x^6}{7}dx\) \(\Rrightarrow\) \(\cfrac {x^7 ln(9x)}{7} - \cfrac {1}{7}\int {x^6}dx\)

\(\Rrightarrow\) \(\cfrac {x^7 ln(9x)}{7} - \cfrac {1}{7} \times \cfrac{x^7}{7}\) \(\Rrightarrow\) \(\cfrac {x^7 ln(9x)}{7} -\cfrac{x^7}{49} + C\)

Hence,

\(x^7 \cfrac {(7ln(9x) -1)}{49} + C\)

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

## [1] TRUE

We can see from above the definite integral of the function is 1 and f(x) is a probability density function.