1 Question 1

Use integration by substitution to solve the integral below. \[\int 4e^{-7x}dx\]

Answer:

Let \(u=-7x\)

then \(du=-7dx\)

then \(\int 4e^{-7x}dx = -\frac{4}{7} \int e^{u}du= -\frac{4}{7}e^u+C=-\frac{4}{7}e^{-7x}+C \;\;\;\;where\;c\;is\;a\;constant\)

2 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:

\(N(t)=\int (-3150t^{-4}-220)dt=\int-3150t^{-4}dt-\int220dt\)

\(=\frac{3150}{3}t^{-3}-220t+C = 1050t^{-3}-220t+C\;\;\;where\;C\;is\;a\;constant\)

\(\because N(1)=6530\)

\(\therefore 1050(1)^{-3}-220(1)+C=6530\)

\(\therefore C=6530+220-1050=5700\)

\(\therefore N(t)=1050t^{-3}-220t+5700\)

3 Question 3

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

Answer:

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

\(=8.5^2-4.5^2-9(8.5)+9(4.5)=16\)

4 Question 4

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

\[y=x^2-2x-2,\;y=x+2\]

Enter your answer below.

Answer:

1. Plot the two functions

eq1 = function(x){x^2-2*x-2}
eq2 = function(x){x+2}

curve(eq1, from=-3, to=5, n=300, xlab="X", ylab="Y", col="red",lwd=2,
      main="Plot the Region Bounded By the Two Equations",)

curve(eq2, from=-3, to=5, n=300, xlab="X", ylab="Y", col="blue",lwd=2, 
      main="Plot the Region Bounded By the Two Equations", add = TRUE)

text(-2.5,10,"y = x^2-2x-2", pos=4, col = 'red')

text(-2.5,-1,"y = x+2", pos=4, col = 'blue')

2. Solve for Intersections

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

Then \(x^2-3x-4=0\)

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

Then \(x_{1}=-1, \;x_{2}=4\)

3. Solve for Area

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

\(=-\frac{1}{3}(4^3-(-1)^3)+\frac{3}{2}(4^2-(-1)^2)+4(4-(-1))\)

\(=20.8333\)

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:

Let c as the inventory cost, s the lot size, n the number of orders, and assume that half of the inventory are in stock,

then \(c = 8.25n+\frac{3.75}{2}s\;while\;ns= 100\)

then \(c = 8.25n +\frac{3.75}{2}\frac{100}{n}=8.25n+206.25n^{-1}\)

To find n that minimize c, let

\({c}'= -206.25n^{-2}+8.25 = 0\)

\(n_{min} = 5\)

then $s_{min} = =22 $ and

\(c_{min}=8.25(5)+206.25(5)^{-1}=82.5\)

6 Question 6

Use integration by parts to solve the integral below.

\[\int ln(9x).x^6dx\]

Answer:

let \(u=ln(9x),\;dv=x^6dx\)

then \(du = \frac{1}{9x}(9)dx=x^{-1}dx,\;v=\frac{1}{7}x^7\)

then \(\int ln(9x).x^6dx = \int udv=uv-\int vdu\)

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

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

\(=\frac{1}{7}ln(9x)x^7-\frac{1}{49}x^7+C\)

\(=\frac{1}{7}x^7(ln(9x)-\frac{1}{7})+C\;where\;C\;is\;a\;constant.\)

7 Question 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.

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

1. \(\because x \in [1,\;e^6]\)

\(\therefore x > 0\)

\(\therefore f(x) >0\)

2. \(\because \int f(x)dx=\frac{1}{6}\int_{1}^{e^6} x^{-1}dx=\frac{1}{6}(ln(e^6)-ln(1))=\frac{1}{6}(6)=1\)

\(\therefore\) all values of (x) sum to 1.

Therefore f(x) is a probability density function.