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

\[\int 4 e^{-7x} \,dx\] We will substitute with \[u=-7x \\ du=-7dx \\ dx = -\frac{1}{7}du\]

As a result \[4 \int e^{-7x} \,dx = -\frac{4}{7} \int e^{u} \,du=-\frac{4}{7} e^{u} + const = -\frac{4}{7} e^{-7x} + const\]

Answer: \[\int 4 e^{-7x} \,dx = -\frac{4}{7} e^{-7x} + const\]

2. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of dN/dt = - 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.

We will find the integral fromthe rate to find the function N(t): \[\int dN(t)\,= \int(- \frac{3150}{t^4}-220)\, dt, \\ N(t)=-\int\frac{3150}{t^4}\,dt-\int220\, dt + c=-\frac{3150}{(-3)t^3}-220t+ c=-\frac{1050}{t^3}-220t+ c\] t=1, N=6530 per problem. By substituting t and N in the function, we will find const: \[6530=-\frac{1050}{1^3}-220 \cdot 1 + c, \\ c = 5700 \] Answer: \[N(t)=-\frac{1050}{t^3}-220t+5700\]

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

The triangle is between 4.5 and 8.5 along the x-axis. We will use integral to find the area of the triangle under line between 4.5 and 8.5: \[A=\int_{4.5}^{8.5} (2x-9 )\,dx=(\frac{2x^2}{2}-9x)|_{4.5}^{8.5}=(8.5^2-9 \cdot 8.5) - (4.5^2-9 \cdot 4.5)=16\] Answer: A=16

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

\[y = x^2 - 2x - 2, y = x + 2\] Let’s plot the graphs of these functions:

x=seq(-5,5,0.1)
y1=x^2 - 2*x - 2
y2=x+2
plot(x,y1,type='l',col='red')
lines(x,y2,col='green')

We will solve equation below to find the x coordinate of the intersection points: \[x^2 - 2x - 2 = x + 2 ,\\ x^2 - 3x - 4=0, \\ (x-4)(x+1)=0, \\ x_1=4, x_2=-1\]

To find the area between x=4 and x=-1, we will find the area under the line and subtract the area under the parabola. To find the area, we will use the integral: \[A=\int_{-1}^{4} (x^2 - 2x - 2)\,dx - \int_{-1}^{4} (x+2)\,dx=(\frac{x^3}{3}-\frac{2x^2}{2}-2x)|_{-1}^{4}-(\frac{x^2}{2}+2x)|_{-1}^{4}=[\frac{4^3}{3}-\frac{2 \cdot 4^2}{2}-2 \cdot4 - \frac{(-1)^3}{3}+\frac{2 \cdot (-1)^2}{2}+2 \cdot (-1)] - [\frac{4^2}{2}+2 \cdot4 - \frac{(-1)^2}{2}+2 \cdot (-1)]=20.833\]

Using R:

f1 <- function(x)
{
  return(x^2 - 2*x - 2)
}


f2 <- function(x)
{
  return(x+2)
}

integrate(f2,lower=-1,upper=4)$value - integrate(f1,lower=-1,upper=4)$value
## [1] 20.83333

Answer: A=20.833

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.

Order cost = 8.25 * 110/x = 907.5/x
Storage cost = 3.75 * x/2 = 1.875x
Cost = 907.5/x + 1.875x,
where x is the size of a lot.
To find minimum cost, we will find the first derivative and make it =0: \[\frac{dC}{dx}=-\frac{907.5}{x^2}+1.875=0, \\ 1.875=\frac{907.5}{x^2}, \\ x_1=22, x_2=-22\] The size can’t be negative, so x=22. The number of orders per year : \[\frac{110}{22}=5\]

Answer: the lot size is 22, the number of orders per year is 5

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

\[\int ln(9x) \cdot x^6\,dx\] The integration by part: \[\int f \, dg=f \cdot g - \int g \, df, \\ f=ln(9x) => df=\frac{1}{x}dx, \\ dg=x^6dx => g=\frac{x^7}{7}\]

The result: \[\int ln(9x) \cdot x^6\,dx = ln(9x) \cdot \frac{x^7}{7} - \int\frac{x^7}{7} \cdot \frac{1}{x}dx + const=ln(9x) \cdot \frac{x^7}{7} - \frac{x^7}{49} + const=\frac{x^7}{49}(7ln(9x)-1)+const\]

Answer: \[\frac{x^7}{49}(7ln(9x)-1)+const\]

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.

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

The function should be non-negative and integral on the interval [1, e^6] should be 1: \[\int_{-1}^{e^6}\frac{1}{6x}\,dx=\frac{ln(x)}{6}|_1^{e^6}=\frac{1}{6}[(ln(e^6)-ln(1)]=\frac{1}{6}(6-0)=1\]

Answer: Both requirements were met, this function is a probability density function on the interval [1, e^6].