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

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

Let u = -7x then du = -7dx

\[4\int { { e }^{ u } } \frac { du }{ -7 }\] \[-\frac { 4 }{ 7 } \int { { e }^{ u } } du\] \[-\frac { 4 }{ 7 } { e }^{ u }+\quad constant\] since u= -7x

\[-\frac { 4 }{ 7 } { e }^{ -7x }+\quad constant\]

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.

\[N'(t)= -\frac{3150}{t^{4}}-220\] \[N(t)\quad =\quad \int { (\frac { -3150 }{ { t }^{ 4 } } -220)dt } \]

when N=1, then

\[N(1)=1050 - 220 + constant = 6530\] \[constant = 5700\]

The function is

\[N(t) = \frac {1050} { t ^{ 4 }} -220t +5700\]

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

eqn <- function(x) {
  2*x-9
}

area <- integrate(f=eqn, lower = 4.5, upper = 8.5)
area
## 16 with absolute error < 1.8e-13

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

y = x2 - 2x - 2, y = x + 2

# define two functions f and g
f <- function(x) {x^2-2*x-2}
g <- function(x) {x+2}

# draw both functions f and g
curve(expr=f, from = -5, to = 5)
curve(expr=g, from = -5, to = 5, add = T)

# find lower intersection
rt <- uniroot(function(x)  f(x) - g(x)  , interval = c(-5,0), tol=0.000001) 
rt$root
## [1] -1
# find upper intersection
rt <- uniroot(function(x)  f(x) - g(x)  , interval = c(0, 5), tol=0.000001) 
rt$root
## [1] 4
# f area between -1 and 1
f_area <- integrate(f, lower = -1, upper = 4)
# g area between -1 and 1
g_area <- integrate(g, lower = -1, upper = 4)

# area in between
g_area$value - f_area$value
## [1] 20.83333

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.

Let x be a number of flat irons to order.

yearly storage cost = storage cost per iron * avg number of irons stored = 3.75 * x/2 = 1.875 * x

yearly order cost = cost of each order * num of orders = 8.25 * 110/x= 907.5/x

inv cost = yearly storage cost + yearly ordering cost = 1.8758 * x + 907.5/x = f(x)

To find the minimized value, we would need differentiate and solve for 0.

f’(x) = 0

\[1.875\quad - \quad \frac { 907.5}{ x^{2} } = 0\]

\[x\quad =\quad \sqrt { \frac { 907.5 }{ 1.875 } } \]

\[ x = 22 \]

so number of orders will be 110/22 = 5 orders per year that will minimize inventory costs.

6. Use integration by parts to solve the integral below.

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

\[ \int { uv'dx\quad =\quad uv\quad -\quad \int { u'vdx } } \] in this case u=ln(9x) v’=\(x^{6}\)

\[ \int { ln(9x).{ x }^{ 6 }dx }\quad = \quad \frac { 1 }{ 7 } ln(9x){ x }^{ 7 }\quad -\quad \frac { 1 }{ 7 } \int { { x }^{ 6 }dx } \quad \]

\[ =\quad\quad \frac { 1 }{ 7 } ln(9x){ x }^{ 7 }\quad -\quad \frac { 1 }{ 7 } (\frac { { x }^{ 7 } }{ 7 } )\quad +\quad constant\quad \quad \] \[ = \frac { { x }^{ 7 } }{ 7 } \left[ ln(9x)\quad -\frac { 1 }{ 7 } \right] \quad +\quad constant\quad \quad \]

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

\[ \int _{ 1 }^{ { e }^{ 6 } }{ \frac { 1 }{ 6x } } dx\quad \quad \] \[ \frac { 1 }{ 6 } \int _{ 1 }^{ { e }^{ 6 } }{ \frac { 1 }{ x } } dx\quad \quad\]

\[ \frac { 1 }{ 6 } (ln({ e }^{ 6 })\quad -ln(1))\quad \] \[\frac { 1 }{ 6 } (6-0)\quad =\quad 1\] The definite integral of the function on interval [1, \(e^{6}\)] is 1. So it satisfies that function is indeed a probability distribution function.