1.

Use integration by substitution to solve the integral below.

\(\int4e^{-7x}dx\)

substitution of u, u=-7x

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

dx = \(-\frac{1}{7}du\)

replace dx, \(4\int-\frac{1}{7}e^udu\)

with substitution of u, \(-\frac{4}{7}e^{-7x}+c\)

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.

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

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

with split and intergration,

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

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

\(N(t) = \frac{1050}{t^3}-220t+c\)

find the day 1 at \(N(t) = 6530\)

\(6530 = \frac{1050}{1^3} - 220(1)+c\)

\(c = 5700\), Therefore, \(N(t) = \frac{1050}{t^3}-220t+5700\)

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

integrate(area, lower = 4.5, upper = 8.5)$value
## [1] 16

4.

Find the area of the region bounded by the graphs of the given equations. \(y = x^2-2x-2, y = x+2\)

with solving x, x = 4 or x = -1

area <- function(x)
{
  x+2 - (x^2-2*x-2)
}

integrate(area, lower = -1, upper = 4)$value
## [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 c=cost,

n=number of orders,

x=number of irons in the order.

then, nx=110, and on average x2 irons in storage.

Therefore, \(c = \frac{x}{2}3.75+8.25n\), while \(nx=110\) \(x=\frac{110}{n}\)

\(c = \frac{110}{n}*\frac{1}{2}*3.75+8.25n\)

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

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

\(n=5\)

The number of orders is 5.

6.

Use integration by parts to solve the integral below. \(\int{ln}(9x)x^6dx\)

\(\int{u}*dv=uv-\int{v}*du\)

Let \(u=ln(9x)\),

then \(du=\frac{1}{x}\), \(dv=x^6\), \(v=\frac{x^7}{7}\)

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

\(ln(9x)*\frac{x^7}{7}-\frac{x^7}{49}+c\)

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.

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

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

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

since \(ln(1)=0\), then \(=\frac{1}{6}6(ln(e))\)

\(=ln(e)=1\)

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