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

x <- Var("x")
sympy("integrate(4 * exp(-7 * x))")

## [1] "-4*exp(-7*x)/7"

2. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \(dN/dt = -3150/t^4 - 220\) 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.

\[ \int - \frac{3150}{t^4}-220 dt = \int -3150t^{-4}-220dt\]

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

\[6350 = \frac{1050}{1^3} - 220(1) + c \] \[= 1050 - 220 + c = 830 + c\]

\[N(t) = \frac{1050}{t^3} - 220t + 5520\]

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

f3 <- function(x) {
  2*x-9
}

area <- integrate(f3, 4.5, 8.5)$value

area

## [1] 16

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

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

\[y1 = y2\] \[x+2 = x^2 - 2x -2 \] \[0= x^2 - 3x-y\] \[0=(x-4)(x+1)\]

interg <- function(x){x + 2 - (x**2 -2*x - 2)}
integrate(interg, -1, 4)

## 20.83333 with absolute error < 2.3e-13

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.

\[n*x = 110\] \[x = 110/n\] \[C'= 8.25 -206.25/n^2\] \[8.25 -206.25/n^2 = 0\]

\[x = \frac{110}{5} = 22\]

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

\[\int ln( 9x ) * x^6 dx\]

x <- Var("x")
sympy("integrate(log( 9 * x ) * x**6)")

## [1] "x**7*log(9)/7 + x**7*log(x)/7 - x**7/49"

7.Determine whether f(x) is a probability density function on the interval [1,e6]. If not, determine the value of the definite integral.

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

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