Homework 13

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

\[ \int{4e}^{-7x}dx \]

\[ u = -7x \] \[ du = -7dx \]

\[ dx = \frac{du}{-7} \] \[ \int{4e^{u} \frac{du}{-7}} \] \[ \frac{4}{-7} \int{e^{u} du}\] \[ \frac{4}{-7}e^{u} + C \]

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.

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

\[ N = - \frac{3150}{3t^3} - 220t + C \] \[ N(1) = - \frac{3150}{3(1)^3} - 220(1) + C = 6530\] \[ -1050 -220 + C = 6530\] \[C = 7800 \]

\[N(t) = - \frac{31503}{t^3} - 220(t) + 7800 \] \[N(0) = - \frac{3150}{3(0)^3} - 220(0) + 7800 \]

\[N(0) = 7800 \]

3. Find the total area of the red rectangles in the figure below, where the equation of the line is

\[f(x) = 2x - 9\] \[\int_{4.5}^{8.5} 2x - 9dx\] \[ = [x^2 - 9x]_{4.5}^{8.5}\] \[= (8.5^2 - 9(8.5)) - (4.5^2 - 9(4.5)) \] \[= 16\]

The area of the shaded red rectangles is 16.

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

\[y = x^2 - 2x -2, y = x + 2 \] Graph for equation:

curve(x^2 - 2*x -2, lwd = 2, xlim= c(-5, 5))
curve(x+2, lwd = 2, xlim= c(-5,5), add=TRUE)

Intersection: \[x^2 - 2x -2 = x+ 2x^2 - 3x - 4 = 0\]

f <- function(a) {
  a^2 - 3*a - 4
}

root <- polyroot(c(-4, -3, 1))

ifelse(Im(root) == 0, Re(root), root)
## [1] -1+0i  4-0i

Area: \[\int_{-1}^{4} x + 2dx - \int_{-1}^{4} x^2 - 2x - 2dx\] \[= -[\frac{1}{3} x^3 - \frac{3}{2}x^2 - 4x]_{-1}^{4}\]

area <- - ( ( ((1/3)*64) - (1.5*16) - 16 ) - (-(1/3) - 1.5 + 4 ) )
area
## [1] 20.83333

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’s say, Orders per year is n,
Lot size is s and Cost is c

Assuming stocking for half of the annual orders.

\[Cost = FixedCost * Orders + (StoringCost*LotSize/2*Orders)\]
\[c=8.25n + \frac{3.75*110}{2*n}\]
\[c=8.25n + \frac{206.25}{n}\]
\[c' = 8.25 - \frac{206.25}{n^2}\]

\[c'=0\]

\[0= 8.25 - \frac{206.25}{n^2}\]
\[n^2 = \frac{206.25}{8.25}\] \[n^2 = 25\]

\[n = 5 \]

Substituting value of n in the above equation

c <- 8.25*5 + (206.25/5)
c
## [1] 82.5

Inventory cost is 82.50 for lot size of 55, with orders per year of 5.

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

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

Choose: \[u = ln(9x), \frac{du}{dx} = x^6\]
\[du = \frac{9}{9x}dx = \frac{1}{x}dx\]
\[dv = x^6 dx\]

\[v = \frac{1}{7} x^7\]

In equation:

\[\int udv = uv - vdu \] \[= ln(9x) \frac{1}{7}x^7 - \frac {1}{7} x^7 \frac {1}{x} dx \]
\[= ln(9x) \frac{x^7}{7} - \frac{x^7}{49} - C \]

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.

\[f(x) = \frac{1}{6x}\]
\[\int _1^{e^6} f(x)dx\]
\[= \int_1^{e^6} \frac{1}{6x} dx \] \[= \frac{1}{6} \int _1^{e^6} \frac {1}{x} dx \] \[= \frac {1}{6} ln(x) _1^{e^6}\]
\[= \frac{1}{6} [ln(e^6) - ln(1)]\] \[= \frac{1}{6}[6 - 0]\] \[= 1\]