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

\[\int4^{-7x}~dx\] \[\\U = -7x\] \[\\~dx = \frac{~dU}{-7}\] \[ \int4^{-7x}~dx = \int\frac{4}{-7}e^U~dU\\ = \frac{-4}{7}e^U + Constant\\ = \frac{-4}{7}e^{-7x} + Constant \]

2. Find a function N(t) to estimate the level of contamination if the level after 1 day was 6530 bacteria per cubic centimeter.

\[ \ N'(t) = \frac{-3150}{t^4} - 220\\ N (t) = \int(\frac{-3150}{t^4} - 220)~dt\\ = \frac{3150}{3t^3} - 220t + constant \] ##### N(1) = 6530 \[ \ N(1) = 1050 - 220 + Constant = 6530\\ \ Constant = 7800 \] ##### The funcsion is \[ \ N(t) = \frac{1050}{t^3} - 220t + 5700 \] ##### 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 = {4.5}^{8.5}(2x - 9)~dx\ = [x^2 - 9x]{4.5}^{8.5}\ = [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.

f1 <- function(x){x^2 - 2*x -2}
f2 <- function(x){x + 2}
curve(f1, -1, 4)
curve(f2, -1, 4, add = TRUE)

$$  Area = {-1}^{4}(x + 2)~dx - {-1}{4}(x2 - 2x -2)~dx\ = [ X^2 + 2X]{-1}^{4} - [X^3 - X^2 - 2x]{-1}^{4}\ = 20

\[ ##### 5. Beauty Supply Store. Let Y be the cost, n be the number of orders per year and x be the number of irons per order x = 110/n \]  Y = 8.25n + 3.75x $$

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

\[ \ \int ln(9x)x^6~dx\\ \int UdV = UV - \int VdU\\ \ U = ln(9x)\\ \ dV = x^6dx \\ \ \int ln(9x)x^6~dx\\ \ = \frac{1}{7}ln(9x)x^7 - \frac{1}{7}\int x^6 dx\\ \ = \frac{1}{7}ln(9x)x^7 - \frac{1}{49}x^7 \] ##### 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. \[ \int_1^{e^6}\frac{1}{6x}~dx\\ \ = [\frac{1}{6}ln(x)]_1^{e^6}\\ \ = \frac{1}{6}(ln(e^6) - ln(1))\\ \ = 1 \]

f(x) is a probability density.