Question 1

Use integration by substitution to solve the integral below.

∫4e−7xdx

Solution Select u to be the g(x) inside f(g(x)) u=−7xdu=−7dx−du/7=dx

Substitute dx with our du (reverse chain rule) pull out the 4 (constant) from the integrand

\ e^{u}+c\ e^{-7x}+c

Question 2.

Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of dN/dt=−3150t(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 are given rate, therfore we need to derive a “position” function by taking the integral of th rate function.

We can re-wite the first term of the integrand using a negative power so that we may easily apply the power rule rather than the quotient rule for integration. Evaluate the integral term by term and then simplify \ \ -3150t^{-4}dt - -220dt\ -3150t^{-4}dt - 220dt\ -3150 t^{-3} - 220t\ N(t)= -220t+c

Now that we have identified our N(t) we need to apply the given initial conditions. We are given that N(1)=6530,hence we want to solve for our constant c using the initial condition

6530= -220(1)+c\ 6530=1050-220+c\ 6530-1050+220=c\ 5700=c

Hence our final answer is

N(t)= -220t+5700

Question 3.

Find the total area of the red rectangles in the figure below, where the 3. equation of the line is f(x)=2x-9 There are 4 rectangles in the plot of this function within the closed interval [4.5,8.5] (It is hard to tell from the image)

We can do this using part 2 of the fundamental theorem of caluclus or by inspection.

By inspection we can calculate the area of each rectangle and then add them up. A=lxw

A1=(1)(5.5−4.5)=1 A2=(1)(6.5−5.5)=3 A3=(1)(7.5−5.5)=5 A4=(1)(8.5−6.5)=7 A=A1+A2+A3+A4=1+3+5+7=16

Using part 2 of the fundamental theorem of calculus

8.5 ∫ 2x−9dx 4.5 Lets use R to evaluate this simple integral at the end points

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


## integrate the function from 0 to infinity
integrate(integrand, lower = 4.5, upper = 8.5)
## 16 with absolute error < 1.8e-13

Question 4 Find the area of the region bounded by the graphs of the given equations. y=x(2)−2x−2y=x+2

fx <- function(x)
  {
  (x^2)-2*x-2
}

gx <- function(x)
  {
  x+2
}

plot (fx, -15, 15)
plot (gx, -5, 5, add=TRUE)

Find the area enclosed by f(x) and by g(x). Finding exact bounds of integration.

Set f(x)=g(x) and solve for x . The intersection of two curves are the bounds of integration. We should get two values for x because of the quadratic term.

x(2)−2x−2=x+2x(2)−3x−4=0(x−4)(x+1)=0x=4x=−1

The formula for finding the area enclosed by two curves is as follows: b ∫ (top−bottom)dx a

As we know a=-1 and b=4,the top function is the function that encloses the upper half of the area and the bottom function is the function that encloses the bottom of the area. In our case, the top function is the straight line while the bottom function is the parabola

{ -1 }^{ 4 }{ (x+2)-(x^{2}-2x-2) } dx\ { -1 }^{ 4 } {(x+2-x^{2}+2x+2)dx}\ _{- 1 }^{ 4 } {(-x^{2}+3x+4)dx}

Using R to integrate and calculate

integrand <- function(x)
  {
  -x^{2}+3*x+4
  }


integrate(integrand, lower = -1, upper = 4)
## 20.83333 with absolute error < 2.3e-13

Question 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 C be cost, n be the number of orders per year and x be the number of irons in an order (lot size).

n∗x=110 x=110n

C=8.25∗n+3.75∗x/2 C=8.25∗n+3.75∗110/n/2 C=8.25∗n+412.5/2n C=8.25∗n+206.25/n C′=8.25−206.25/n(2)

Determine the value of n by equating this expression to 0. 8.25−206.25/n(2)=0 n(2)=206.25/8.25 n=√25 =5 x=110/5=22

##  lot size x = 22 and the number of orders per year n = 5

Question 6

Use integration by parts to solve the integral below ∫ln(9x)x(6)dx formula for integration by parts: We pick dv to be the term that can be integrated easily while u is the term that can be derived easily.

uv−∫vdu u=ln(9x) du=1/xdx dv=x(6) v=1/7x(7)

we have the following integral, let us evaluate

x7ln(9x)/7−∫1/7(x(7)1/x)dx x7ln(9x)/7−1/7∫(x(7)/x)dx x7ln(9x)/7−1/7∫x6dx x(7)ln(9x)/7−1/7(x(7)/7)+C x(7)ln(9x)/7−x(7)/49+C

x(7)ln(9x)/7−x(7)/49+C x(7)/7[ln(9x)−1/7]+C

Question 7

Determine whether f(x) is a probability density function on the interval 1, e6 . If not, determine the value of the definite integral f(x)=1/6x

e6 ∫1/6xdx

e6 1/6∫1/xdx 1 1/6(ln(e(6)−ln(1)) 1/6(6−0)=1

f(x) is a probability density function on the closed interval [1,e6]