WebTest Name: (Test)

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

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

Solution

let -7x=u

-7dx=du

dx= -(1/7)du

\[\begin{equation} \begin{split} \int 4e^{-7x} \mathrm{d}x &= \int 4e^u \times -(1/7) \mathrm{d}u\\ &= -\frac{4}{7}\int e^u \mathrm{d}u \end{split} \end{equation}\]

The integral of \(e^u\) with respect to u is \(e^u\).

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

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

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.

Solution

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

\[\begin{equation} \begin{split} \int (-\frac{3150}{t^4}-220) \mathrm{d}t &= -1 \int \limits_0^1 (\frac{3150}{t^4}+220) \mathrm{d}t\\ &= (-1(-\frac{3150}{3}t^{-3}+220t) +C) \Big|_0^1 \\ &= (\frac{3150}{3}t^{-3}-220t +C) \Big|_0^1 \end{split} \end{equation}\] \[\begin{equation} (\frac{3150}{3}t^{-3}-220t + C) \Big|_0^1 = 6530 \end{equation}\] \[\begin{equation} N(t) = \frac{3150}{3}t^{-3}-220t + 5700 \end{equation}\]

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

!(https://raw.githubusercontent.com/YunMai-SPS/DATA605_homework/master/data605_week13/DA605_week13_3.png)

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

curve(func,col = 'blue', lwd = 2, from = 0, n = 15, xlim=c(-5, 15), ylab='f(x)')
abline(v=c(4.5,8.5), col=c("red", "red"))

Solution

\[\begin{equation} \begin{split} \int\limits_{4.5}^{8.5} (2x-9) \mathrm{d}t &= x^2 -9x +C\Big|_{4.5}^{8.5}\\ &= 8.5^2-4.5^2-9 (8.5-4.5)\\ &= 16 \end{split} \end{equation}\]

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.

Solution

func1 <- function(x){x^2 - 2*x - 2}
func2 <- function(x){x + 2}
curve(func1, col = 'blue', lwd = 2, from = -5, n = 100, xlim=c(-5, 5), ylab='f(x)')
curve(func2, col = 'blue', lwd = 2, from = -5, n = 100, xlim=c(-5, 5), ylab='f(x)',add=TRUE)

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

# zeros of f
root <- polyroot(c(-4, -3,1))
ifelse(Im(root) == 0, Re(root), root)
## [1] -1+0i  4-0i
\[\begin{equation} \begin{split} \int\limits_{-1}^{4}(x + 2)-(x^2 - 2x - 2) \mathrm{d}t &= \int\limits_{-1}^{4}-x^2+3x+4 \mathrm{d}t &= -\frac{1}{3}x^3+\frac{3}{2}x^2+4x+C\Big|_{-1}^{4}\\ &= 20.83333 \end{split} \end{equation}\]

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.

Assuming the stock could be replenished just as they become empty and we order n times every year, there are 110/n flat irons to begin with. In every restocking period, there is (110/n - 110t) left in the shelf. So the cost per each restocking period is \(\int _0^{\frac{1}{n}} 3.75*(110/n - 110t)\).

The cost per year f(t) is:

\[\begin{equation} \begin{split} f(t) &= (\int _0^{\frac{1}{n}} 3.75 \times(\frac{110}{n} - 110t)\mathrm{d}t+8.25) \times n\\ &= 3.75n(\frac{110}{n}t-\frac{110}{2}\times t^2)\Big|_0^{\frac{1}{n}}+8.25n\\ &= \frac{206.25}{n}+8.25n \end{split} \end{equation}\]

now we have c(n)= +8.25n. Set c’(n)=0, then

f=expression(206.25/n+8.25*n)
D(f,'n')
## 8.25 - 206.25/n^2

set 8.25 - 206.25/n^2 = 0 , then

n = 5

So keeping the lot size as 22 and order times per year as 5 will minimize the inventory costs.

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

Solution

\(\int ln(9x) · x^6dx\)

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

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

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

\[\begin{equation} \begin{split} \int u dv &= uv- \int v du\\ &= ln(9x)\frac{1}{7}x^7 -\int \frac{1}{7}x^7 \frac{1}{x}dx\\ &=ln(9x)\frac{x^7}{7} - \frac{x^7}{49}-C \end{split} \end{equation}\]

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 ) = 1/6x

Solution

\[\begin{equation} \begin{split} \int_1^{e^6} 1/6x &= (log(x)/6+C)\Big|_1^{e^6}\\ &= (log(e^6)-log(1))/6\\ &=1 \end{split} \end{equation}\]

So f ( x ) is a probability density function on the interval \([1, e^6]\).