Problem 1

Use integration by substitution to solve the integral below.

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

Let \(u=-7x\), then \(du = -7dx\). \(dx = \frac {-du}7\)

\[ 4\int e^U\frac{dU}{-7} \\ \frac{4}{-7}\int e^UdU \\ \frac{4}{-7} e^U + C \\ \frac{4}{-7} e^{-7x} + C \]

Problem 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.

\[\frac {dN} {dt} = \frac {- 3150} {t^4} - 220 \] \[N = 3150 \frac {t^-3} {-3} - 220t + C \] t=1 N=6530

\[6530 = -1050 (1^{-3}) - 220(1) + C \] \[6530 + 1270 = C \] \[7800 = C \]

\[N = \frac {-3150}{3t^3} - 220t + 7800 \] \[N = \frac {-1050}{t^3} - 220t + 7800 \]

Problem 3

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

Area of 1st rectangle = 1 Asrea of 2nd rect = 3 Area of 3rd rect = 5 area of 4th rect = 7 total area = 1 +3+5+7 = 16

using integral \[ Area = \int_{4.5}^{8.5} 2x - 9 dx\]

Area <- function(x) {2*(x)-9}
integrate(Area, lower = 4.5, upper = 8.5)
## 16 with absolute error < 1.8e-13

Problem 4

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

\(y_1 = x_1^2 - 2x_1-2\) \(y_2 = x_2 + 2\)

by solving above equations we get values -2 and 5 for x

f1 <- function(x) x^2-2*x-2
f2 <- function(x) x+2

x1 <- seq(-2, 5, .5)
plot(x1, f1(x1), type='l', col="red", 
     xlab="", ylab="")
lines(x1, f2(x1), col="blue")
abline(h=0)

From the graph we see that Area = area of f2 - area of f1 and the points we notice here is -1 and 4 so by integrating we can find the area

A1 <- function(x) {(x)+2}
integrate(A1, lower = -1, upper = 4)
## 17.5 with absolute error < 1.9e-13
A2 <- function(x) {(x)^2 - 2*(x) -2}
integrate(A2,lower = -1, upper = 4)
## -3.333333 with absolute error < 1.2e-13

So total area = 17.5 - (-3.33) = 20.8

Problem 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 \(x\) be a number of flat irons to order.

Yearly storage cost = Storage cost per iron} * Average number of irons store} = 3.75 * x/2 = 1.875x

Yearly ordering cost = Cost of each order * Number of orders = 8.25 * 110/x = 907.5/x

Inventory cost = Yearly storage cost + Yearly ordering cost = 1.875x+907.5/x = f(x)

To find the minimized value, differentiate and solve at \(0\):

\[ \begin{split} f'(x) &= 1.875-\frac{907.5}{x^2} \\ f'(x) &= 0 \\ 1.875-\frac{907.5}{x^2} &= 0 \\ 1.875&= \frac{907.5}{x^2} \\ 1.875x^2&= 907.5 \\ x^2&= \frac{907.5}{1.875} \\ x&= \sqrt{\frac{907.5}{1.875}} \\ x&=\sqrt{484} \\ x&=22 \end{split} \]

Each order should contain \(22\) flat irons, so there should be \(110/22=5\) orders.

Problem 6

Use integration by parts to solve the integral below.

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

Let \(u= ln(9x)\), then \(\frac{du}{dx}=\frac{1}{x}\).

Let \(\frac{dv}{dx}=x^6\), then \(v = \int{x^6 dx} = \frac{1}{7}x^7\).

Using the formula for integration by parts: \(\int{u \frac{dv}{dx}dx} = uv - \int{v \frac{du}{dx} dx}\)

\[ \begin{split} \int{ln(9x) \times x^6 dx} &= \frac{1}{7}x^7 \times ln(9x) - \int{\frac{1}{7}x^7 \times \frac{1}{x} dx} \\ &=\frac{1}{7}x^7 \times ln(9x) - \int{\frac{1}{7}x^6 dx} \\ &=\frac{1}{7}x^7 \times ln(9x) - \frac{1}{7}x^7 \\ &=\frac{1}{7}x^7 (ln(9x) - \frac{1}{7})\\ \end{split} \]

Problem 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.

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

\[ f(x) = \frac {1} {6x} \] \[ F(x) = \int_{1}^{e^6} f(x) dx = 1 \] substituting f(x)

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

\[ F(x) = \frac {1} {6} \bigg[ln(x) \bigg]_{1}^{e^6} \]

\[ F(x) = \frac {1} {6} \bigg[ln(e^6) - ln(1) \bigg] \] \[ F(x) = \frac {1} {6} \bigg[6 - 0 \bigg] = 1 \]

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