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

$$4e^{-7x} , dx = 4 e^{-7x} , dx \ = 4 (e^{-7x}/-7)+C = -4/7, e^{-7x} + C

$$

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.

\[ \int dN/dt \, dt = \int (- 3150/t^4 - 220)\, dt \]

\[N(t) = -(-3150 t^{-3}/3 - 220t + C) \] In order to find C, we substitute the value of N(t) after day 1, which is provided as 6530:

\[6530= -(-3150 (1)^{-3}/3 - 220(1) + C) \]

\[c = 6530 + 220 - 3150/3 = 5700 \] So the final function for the contamination can be stated as follows:

\[N(t) = 1050 t^{-3} - 220t + 5700 \] ### 3. Find the total area of the red rectangles in the figure below, where the equation of the line is f(x) = 2x - 9.

Since the width of each rectangle is 1, we can multiply it by the total length of the 4 rectangles, to find the total area of the red rectangles. This is indicated by: A = [f(5)+f(6)+f(7)+f(8)] x 1 A = [1 + 3 + 5 + 7] = 16

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

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

The first equation is quadratic indicating that it represents a curve (parabola), and the second equation is linear, indicating a straight line. In order to visualize the region bounded by the curve and the line, we assume some arbitary range for x and compute y based on the given equations.

x1 = c(-8:8)
y1 = x1^2-2*x1-2
x2 = c(-8:8)
y2 = x2+2
plot(x1,y1,type="l", xlab="x1,x2", ylab="y1, y2")
lines(x2,y2,type="l")

We see that the curve and the line intersect in 2 points. To find the co-ordinates of these 2 points, we solve by setting the 2 equations equal to each other:

\[ x^2 - 2x - 2 = x + 2\] \[ x^2 - 3x - 4 = 0 \\ = (x - 4) (x + 1)= 0\\ Therefore\, x=4 \,and \, x=-1\]

Solving for y by back-substitution, we get y = 6 when x = 4, and y = 1 when x = -1. So the points of intersection are (4,6) and (-1,1). In order to find the area bounded by the parabola and the line, we integrate the functions between these 2 intersection points, as shown below:

\[ \int_{-1}^{4} (x + 2) - (x^2 - 2x - 2)\, dx \\ = \int_{-1}^{4} -x^2 + 3x + 4 \, dx \\ = |(-x^3/3 + 3/2x^2 + 4x)|_{-1}^{4} \\ = 20.83 \] The area bounded by the curve and the line is 20.83 sq units. That is confirmed via the R function below:

area <- function(x) (-x^2+3*x+4) 
integrate(area,4,-1)
## -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.

Let N = number of irons per order Then number of orders O = 110/N Assuming an even rate of sales throughout the year, we get Storage cost S = 3.75 x N/2 Total Fixed cost F = 8.25 x O = 8.25 x 110/N

Total inventory cost I = S + F I = 3.75xN/2 + 8.25x110/N I = 1.875N + (8.25*110)/N

(8.25*110)
## [1] 907.5

I = 1.875N+907.5/N To minimize this cost function, we set its first derivative to 0 dI/dN = 0 1.875-907.5/N^2 = 0 N^2 = 907.5/1.875

(N=sqrt(907.5/1.875))
## [1] 22

Back-substituting we get, O = 110/22 = 5

Therefore to minimize inventory costs, the lot size and number of orders should be 22 and 5 respectively

6. Use integration by parts to solve the integral below: ln(9x)*x^6 dx

\[ Set\, u = ln(9x)\, and \,dv = x^6 dx \\ Then\, du = dx/x\, and\, v = x^7/7 \] Using integration by parts, we get \[ \int udv = uv - \int vdu \\ \int ln(9x).x^6dx = ln(9x).(x^7/7) - \int (x^7/7).(1/x)dx\\ = ln(9x).(x^7/7) - (1/7).(x^7/7)\\ = ln(9x).(x^7/7) - (x^7/49) \]

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) = 1/6x

In order for f(x) to be a valid probability density function on the given interval, it needs to satisfy 2 conditions over that interval: 1) f(x)>=0 because probability cannot be negative 2) \[ \int f(x) = 1\] because sum of all probabilities must add up to 1

exp(6)
## [1] 403.4288

Given that f(x)= 1/6x, and e^6 is a positive quantity, it can be safely assumed that the probability will not be negative in the inteval 1 and e^6

Next we evaluate the second condition below: \[ \int_{1}^{e^6} 1/6x\, dx = 1/6.ln(x)|_{1}^{e^6}\, = 1/6[ln(e^6) - ln(1)] \]

(1/6*(log(exp(6)-log(1))))
## [1] 1

This confirms that both the conditions are true and this function is a valid probability density function over this interval.