Data 605 Assignment 13

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

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

\(u = -7x\)

\(du = -7dx\)

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

\(\int 4e^{u}\frac{du}{-7}\)

\(\frac{4}{-7}\int e^udu\)

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

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

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.

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

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

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

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

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

\(N(1) = - \frac{3150}{3(1)^3} - 220(1) + C = 6530\)

\(C = 7800\)

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

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

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

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

\(= [x^2 - 9x]|_{4.5}^{8.5}\)

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

\(= 16\)

Area of 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 point: x^2 -2x-2 = x+2 x^2 -3x-4 = 0

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


# zeros of f 
root <- polyroot(c(-4, -3, 1)) 
ifelse(Im(root) == 0, Re(root), root)
## [1] -1+0i  4-0i

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

\(=-[\frac{1}{3}x^3 - \frac{3}{2}x^2 -4x]|_{-1}^{4}\)

\(=20.8333\)

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.

Number of orders per year = n lot size = s Cost = c

ns = 110 s = 110 / n

Assume half of inventory keep in stocks:

\(c = 8.25n + \frac{375}{2n}\)

\(c = 8.25n + \frac{206.25}{n}\)

\(c' = 8.25 - \frac{206.25}{n^{_{2}}}\)

\(c'=0\)

\(0 = 8.25 - \frac{206.25}{n^{_{2}}}\)

\(n = 5\)

Number of order per year is 5

Lot size is 22 and inventory cost is 78.75.

6. Use integration by parts to solve the integral below

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

Choose:

\(u=ln(9x)\), \(\frac{dv}{dx}=x^{6}\)

\(du = \frac{9}{9x}dx = \frac{1}{x}dx\)

\(dv = x^6dx\)

\(v = \frac{1}{7}x^7\)

In equation:

\(\int u dv= uv-\int v du\)

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

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

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.

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

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

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

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

\(= \frac{1}{6} ln(x)|_1^{e^6}\)

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

\(=\frac{1}{6} [6 - 0]\)

\(=1\)