1. Use integration by substitution to solve the integral below. \(\int4e^{-7x}dx\)

Let \(u=-7x\)

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

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

Now

\(\int4e^{-7x}dx=\int4e^u*-\frac{1}{7}du=-\frac{4}{7}\int e^u*du=-\frac{4}{7}*e^u+C=-\frac{4}{7}*e^{7x}+C\)

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.

Let’s integrate \(\frac{dN}{dt}\)

\(\int-\frac{3150}{t^4}-220 = -\frac{3150}{t^3}*\frac{1}{3}-220*t+C=-\frac{1050}{t^3}-220*t+C\)

Let’s find C when t=1 and level of contamination equals to 6530.

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

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

\(-1270+C=6530\)

\(C=6530+1270=7800\)

The function N(t) to estimate the level of contamination is

\(N(t)=-\frac{1050}{t^3}-220*t+7800\)

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

Let’s integrate the function f(x) from the left limit of 4.5 to the right limit of 8.5 of the rectangles

\(\int_{4.5}^{8.5}2x-9=\frac{2*8.5^2}{2}-9-(\frac{2*4.5^2}{2}-9)=72.5-9-20.25+9=52.25\)

The total area of the red rectangles equals to 52.25

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

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

Let’s set \(x^2 - 2x=x + 2\) to find their points of integration

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

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

Since a=1 \(x_1=\frac{-b+sqrt(b^2-4ac)}{2a}=\frac{-3+sqrt(b^2-4ac)}{2a}\) \(x_2=\frac{-b-sqrt(b^2-4ac)}{2a}\)

a <- 1
b <- -3
c <- -2
x1 <- (-b+sqrt(b^2-4*a*c))/2*a
x2 <- (-b-sqrt(b^2-4*a*c))/2*a

x1
## [1] 3.561553
x2
## [1] -0.5615528

Now, let’s integrate the function \([x + 2-(x^2 - 2x)]dx\) from the left limit of x2 to the right limit of x1

\(\int_{x2}^{x1}[x + 2-(x^2 - 2x)]dx=\int_{x2}^{x1}[x + 2-x^2 + 2x)]dx=\int_{x2}^{x1}[-x^2 + 3x +2)]dx\)

area <- (-x2^3)/3 + 6*x2/2 +2 - (-x1^3)/3 + 6*x1/2 +2
area
## [1] 28.11805

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 the number of orders per year. Then \(frac{110}/{x}\) is

The cost function of orders per year is described by the following formula:

\(f(x)=\frac{110}{x}*3.75 + 8.25*x=\frac{412.5}{x}+ 8.25*x\)

Let’s find critical points.

\(f'(x)= \frac{33}{4}-\frac{825}{2x^2}\)

\(\frac{33}{4}-\frac{825}{2x^2}=0\)

\(\frac{825}{2x^2}=\frac{33}{4}\)

\(x^2=50\)

\(x_1=\sqrt{50}\), \(x_2=-\sqrt{50}\) (let’s ignore \(x_2\) as it’s negative)

In order to calculate minimum, let’s find the second derivative of the function.

\(f''(x)= \frac{825}{x^3}\)

\(f''(\sqrt{50})= \frac{825}{\sqrt{50}^3}\)

round(825/(sqrt(50)^3),0)
## [1] 2

The optimal number of orders per year is 2. The number of irons in the order is \(\frac{110}{2}=55\)

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

 \(\int ln( 9x ) *x^6 dx\)

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

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

So,

\(\int ln( 9x ) *x^6 dx=ln( 9x )*\frac{x^7}{7} - \int \frac{x^7}{7} * \frac{9}{x} = ln( 9x )*\frac{x^7}{7} - \int \frac{9*x^6}{7} = ln( 9x )*\frac{x^7}{7} - \frac{9*x^7}{49}=\frac{1}{7}*x^7(ln(9x)-\frac{9}{7})\)

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) is a probability density function for x on \([1, e^6]\) if

  1. \(f(x) ≥ 0\) for all x in \([1, e^6]\)
  2. \(\int_{1}^{e^6}f(x)dx = 1\)

Let’s check the first condition.

# prepare an input data
x <- seq(1, exp(1)^6, by = 0.01)
y <- 1/(6*x)

#draw graph
plot(y)

The graph proves that f(x) is positive for all x in \([1, e^6]\)

Let’s check the second condition

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

int_f <- (log(abs(exp(1)^6)))/6-(log(abs(1)))/6
int_f
## [1] 1

f(x) is a probability density function for x on \([1, e^6]\) since both conditions are met.