Use integration by substitution to solve the integral below

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

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

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

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

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

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

2.-Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of

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

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.

\(N'(t)=\frac{dn}{dt}=\int(-\frac{3150}{t^4}-220)dt\)

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

\(N(1)=1050-220+C=6530\)

\(C=5700\)

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

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

\(\int^{8.5}_{4.4}(2x-9)dx\)

\(x^2-9xdx|^{8.5}_{4.5}\)

\(\therefore ((8.5)^2-9*(8.5) - (4.5)^2-9*(4.5) = 16\)

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

eqn1 <- function(x){
  return(x**2 -2*x - 2)
}

eqn2 <- function(x){
  return(x + 2)
}

x <- seq(-2, 5, 1)
y <- eqn2(x)

plot(eqn1, from = -2, to = 5, line = "l", col = 'blue')
## Warning in plot.window(...): "line" is not a graphical parameter
## Warning in plot.xy(xy, type, ...): "line" is not a graphical parameter
## Warning in axis(side = side, at = at, labels = labels, ...): NAs introduced
## by coercion

## Warning in axis(side = side, at = at, labels = labels, ...): NAs introduced
## by coercion
## Warning in box(...): "line" is not a graphical parameter
## Warning in title(...): NAs introduced by coercion
lines(x,y, col = 'green')

\(y=x^2-2x-2, y=x+2\)

\(x^2-2x-2=x+2\)

\(x^2-3x-4=0\)

\(x=1, x=-4\)

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

\(\int_{-1}^{4}(-x^2+3x+4)dx\)

\(-\frac{x^3}{3}+\frac{3x^2}{2}+4x|^{4}_{-1}\)

\(\therefore (-\frac{4^3}{3}+\frac{3(4)^2}{2}+4(4))-(-\frac{-1^3}{3}+\frac{3(-1)^2}{2}+4(-1))=20.83333\)

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 \(x\) be a number of flat irons,

Yearly storage cost = Storage cost per iron × Average number of irons =\(3.75.\frac{x}{2}=1.875x\)

Yearly ordering cost = Cost of each order × Number of orders = \(8.25.\frac{110}{x}=\frac{907.5}{x}\)

\(lot.size=x=22, order=\frac{110}{22}=5\)

The lot size is 22 and the number of orders per year of 5 will minimize inventory costs.

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

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

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

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

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

\(\int ln(9x).x^6 dx=\frac{1}{7}x^7.ln(9x)-\int \frac{1}{7}x^7.\frac{1}{x}dx\)

\(\frac{1}{7}x^7.ln(9x)-\int \frac{1}{7}x^6dx\)

\(\frac{1}{49}(7ln(9x)-1)+C\)

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.

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

\(\int^{e^6}_{1}\frac{1}{6x}dx=\frac{1}{6}ln(x)|^{e^6}_{1}\)

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

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