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

\(\int 4 e^{-7x}dx\)

Solution -

let u = -7x

du/dx = -7

dx = du/-7

By substitution -

\(\int 4 e^{-7x}dx\) = 4\(\int e^u du/-7\) = \(-4/7\int e^udu\) = -4/7 * e\(^u\) + 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.

Solution -

First find the general function for N(t) -

\(\int-3150/t^4-220\, dt\) = -3150 * (t\(^-3\)/-3) - 220t + C = N

After 1 day, N = 6530, solving for C

C = 6530 + 3150 * (1^-3)/-3 + 220*(1)
C
## [1] 5700

N(t) = -3150 * (t\(^{-3}\)/-3) - 220t + 5700 = 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.

A = (5.5 - 4.5) * (1 - 0) + (6.5 - 5.5) * (3 - 0) + (7.5 - 6.5) * (5 - 0) +  (7.5 - 6.5) * (7 - 0)
A
## [1] 16

4. Find the area of the region bounded by the graphs of the given equations. y = x\(^2\) - 2x - 2, y = x + 2

x1 = c(-5:5)
y1 = x1^2 - 2*x1 - 2
x2 = c(-5:5)
y2 = x2 + 2
plot(x1,y1,type="l", xlab="x1,x2", ylab="y1, y2")
lines(x2,y2,type="l")

Find the Intersection points of the 2 equations -

y = x\(^2\) - 2x - 2
y = x + 2

Solving for x -

x + 2 = x\(^2\) - 2x - 2
0 = x\(^2\) - 3x - 4
= (x -4) (x + 1)

x = 5, -1

Solving for y - y = x + 2

y = 7, 1

Instersection points are - (5, 7) and (-1, 1)

Integrate the difference between the top function (line) and the bottom function (parabola) between the 2 intersection points in the graph.

\(\int_{-1}^{5} (x+2) - (x^2 - 2x - 2)dx\) = \(\int_{-1}^{5} -x^2+3x+4\)
\(-x^3/3 = 3x2/2 + 4x\), from x =5 to x=-1

-(5)^3/3 + (3*(5)^2)/2+ 4*(5) - (-(-1)^3/3 + (3*(-1)^2)/2+ 4*(-1))
## [1] 18


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.

Solution -

Let -
x - number of orders/year
y - lot size per order

then x * y - total number of irons ordered for the year

x*y >= 110, valid values for x are > = 1 (since we need a minimum order of 1 per year to ship the inventory)

Cost of inventory per year = 3.75 (yx - 110) + 8.75x = 3.75yx + 8.75*x - 412.5

Since our objective is to keep our cost of inventory per year at the minimum, we need to minimize x and maximize y. From the cost fucntion above, the mimumum value for x is 1 and since y = 110/x, y = 110.

By substitution, minimum cost is -

C = 3.75 * ((110)*(1) - 110) + 8.75*(1)
C
## [1] 8.75

x - number of orders/year = 1
y - lot size per order = 110

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

\(\int ln(9x) x^6 dx\)

Solution -
using integration by parts -

\(\int u*dv/dx*dx\) = u*v - \(\int v*du/dx*dx\)

u = ln(9x)

du/dx, using chain rule -

u = 9x d/du[ln u * d/dx(9x)] = 1/u* d/dx (9X) = 1/9x * 9
du/dx = 1/x

dv/dx = \(x^6\)
v = \(x^7\)/7

= ln (9x) * \(x^7\)/7 - \(\int x^7/7 * 1/x* dx\) = \(x^7\)*(ln (9x)/7 - 1)

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

Solution -

\(\int_{1}^{e^6} 1/(6x) dx\)

1/6 * ln x from [1, e\(^6\)], =

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

f (x) is a probability density function on the interval [1, e\(^6\)] - with the area of the curve as 1 (100% probabily for the interval)