Question 1.

Use integration by substitution to solve the integral below.

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

Answer 1.

\(u=-7x\)

\(\frac{du}{dx}=-7\)

\(du=-7dx\)

\(dx=\frac{du}{-7}=\frac{1}{7}du\)

\(\frac{-4}{7}\left(\int e^{u}\; dx\right)\)

\(\frac{4}{7}e^{u}\)

\(\frac{4}{7}e^{-7x}+C\)

Question 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.

Answer 2.

Function of N(t)–> after 1 day level 6530 bacteria per cubic centimer.

\(\frac{dN}{dt}=-\frac{3150}{t^{4}}-220\)

\(dN=-\frac{3150}{t^4}-220dt\)

\(dN=\int -\frac{3150}{t^{4}}\;dt - \int 220;dt)\)

\(N(t)=\frac{3150}{3^t{3}}-220t +C\)

\(N(1)=\frac{3150}{3}-220 + C = 6350\)

\(C=830\)

\(N(t)=\frac{3150}{3t^{3}}-220t+830\)

Question 3.

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

Answer 3.

\(f(x)=2x+9\)

\(\int_{4.5}^{8.5}2x-9x dx\)

\(x^2-9xdx|_{4.5}^{8.5}\)

\(=((8.5)^{2}-9(8.5))-(4.5^{2}-9(4.5))\)

\(=16\)

Question 4.

Find the area of the region bounded by the graphs of the given equations. \(y = x2 - 2x - 2\) , \(y = x + 2\)

Answer 4.

\(f_1(x)=x2-2x-2\)

\(f_2(x)=y = x + 2\)

# create both functions

f_1 <- function(x){
  x^2-2*x-2
}

f_2 <- function(x){
  x+2
}

#plot both functions

curve(f_1, -10, 10)
plot(f_2, -10, 10, add=TRUE)

Find the intersection points.

\(x^{2}-2x-2=x+2\)

\(x^{2}-3x-4=0\)

\((x-4)(x+1)=0\)

x=4 and x=-1 are the intersection points.

The area between these two functions

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

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

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

=21.5

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.

Answer 5.

x is number of orders per year

y is the lot size per order

\(t=x*y\) is the total number of irons ordered per year

To minimize the cost we need to we can keep x=1 and y=110 and t=110

c <- 3.75*(110-110)+(1*8.25)
c
## [1] 8.25

Question 6

Use integration by parts to solve the integral below.

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

Answer 6

Using Integration by Parts:

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

\(u=ln(9x)\)

\(du/dx=1/x\)

\(dv/dx=x^{6}\)

$v=x^{7}/7

$ ln(u)*(v)=uv- v dv$

\(=(ln(9x)x^{7})/7 - (1/7) \int x^{6} dx\)

\(=(x^{7}/7)ln(9x)-(1/7)\)

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

Answer 7

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

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

\(\frac{1}{6}(ln(e^{6})-ln(1))\)

\(\frac(1){6}(6*1-0)\)

\(=1\)