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

\[ \int { 4{ e }^{ -7x }dx } \] Selecting \(u\) to be the \(g(x)\) inside \(f(g(x))\)

\[ u=-7x\\ du=-7dx\\ -\frac{du}{7}=dx \]

Substitute \(dx\) with \(du\) (reverse chain rule)

\[ \frac{-4}{7} \int { { e }^{ u }du } \\ = \frac{-4}{7}e^{u}+C\\ = \frac{-4}{7}e^{-7x}+C \]

Answer: \(\frac{-4}{7}e^{-7x}+C\)

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.

\[ \int {(\frac{-3150}{t^{4}}-220)dt}\\ \int {(-3150{t^{-4}}-220)dt}\\ \int-3150t^{-4}dt - \int -220dt\\ -3150\int t^{-4}dt - 220\int dt\\ -3150 \frac{-1}{3} t^{-3} - 220t\\ N(t)= \frac{1050}{t^{3}}-220t+c \]

\[ \frac{dN}{dt} = N'(t) = \frac{-3150}{t^4}-220 \\ \int{(\frac{-3150}{t^4}-220) dt} = \frac{1050}{t^3}-220t+C = N(t) \]

\[ \begin{split} N(t) &= \frac{1050}{t^3}-220t+C \\ N(1) &= 6530 \\ \frac{1050}{1^3}-220\times 1 +C &= 6530 \\ C &= 6530 - 1050 + 220 \\ C &= 5700 \\ &N(t) = \frac{1050}{t^3}-220t+5700 \end{split} \]

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

3. Find the total area of the red rectangles in the figure below, where the equation of the line is

\[ f(x)=2x-9 \]

knitr::include_graphics("/Users/mohamedhassan/Downloads/hw13_image.png")

When looking at the figure, the red rectangles on the x-axis begin at 4.5 and end at 8.5.

\(Area = \int_{4.5}^{8.5}{(2x-9)dx} = 16\)

\[ \begin{split} \int_{4.5}^{8.5}{(2x-9)dx} \\ = x^2 - 9x|_{4.5}^{8.5} \\ Area = (8.5^2−9(8.5))−(4.5^2−9(4.5)) \\ A = (72.25−76.5)−(20.25−40.5) \\ A = (−4.25)−(−20.25) \\ A = 16 \end{split} \]

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

\[y = x^2 - 2x - 2, y = x + 2\]

equation1 <- function(x) x^2-2*x-2
equation2 <- function(x) x+2

min <- -2
max <- 5
x1 <- seq(min, max, 0.05)
plot(x1, equation1(x1), type='l', col="red", 
     xlab="", ylab="")
lines(x1, equation2(x1), col="blue")
abline(h=0)

Roots of quadratic function \(f_1(x)\):

root <- polyroot(c(-2, -2, 1))
root
## [1] -0.7320508+0i  2.7320508-0i

Finding intersection between each function where \(f_1(x)-f_2(x)=0\):

\[ (x^2-2x-2)-(x+2)=0 \\ x^2-3x-4=0 \] Finding root of \(x^2-3x-4\):

intersection <- polyroot(c(-4, -3, 1))
intersection
## [1] -1+0i  4-0i

The intersection occurs where \(x =-1\) and \(x = 4\)

Plotting the four points we calculated using the polyroot function:

four_points <- c(intersection[1], root, intersection[2])
four_points
## [1] -1.0000000+0i -0.7320508+0i  2.7320508-0i  4.0000000-0i
plot(x1, equation1(x1), type='l', col="red", 
     xlab="", ylab="")
lines(x1, equation2(x1), col="blue")
points(four_points, equation1(four_points))
text(four_points, equation1(four_points), labels = c("a","b","c","d"), pos = 3)
abline(h=0)

a1 <- integrate(equation2, lower=four_points[1], upper=four_points[4])
a2 <- integrate(equation1, lower=four_points[1], upper=four_points[2])
a3 <- integrate(equation1, lower=four_points[3], upper=four_points[4])
a4 <- integrate(equation1, lower=four_points[2], upper=four_points[3])
a1
## 17.5 with absolute error < 1.9e-13
a2
## 0.1307683 with absolute error < 1.5e-15
a3
## 3.464102 with absolute error < 3.8e-14
a4
## -6.928203 with absolute error < 7.7e-14
# Combining the results
area <- a1$value-a2$value-a3$value-a4$value
area
## [1] 20.83333

Area of the region bounded by the graphs of the given equations: 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.

\[ \begin{split} f'(x) &= 1.875-\frac{907.5}{x^2} \\ f'(x) &= 0 \\ 1.875-\frac{907.5}{x^2} &= 0 \\ 1.875&= \frac{907.5}{x^2} \\ 1.875x^2&= 907.5 \\ x^2&= \frac{907.5}{1.875} \\ x&= \sqrt{\frac{907.5}{1.875}} \\ x&=\sqrt{484} \\ x&=22 \\ &110/22 = 5 \end{split} \]

To sell 110 flat irons during the next year, the lot size is 22 and the number of orders per year is 5.

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

\[\int{ln(9x) \times x^6 dx}\]

\[ uv-\int vdu \\ u=ln(9x)\\ du=\frac{1}{x}dx\\ dv=x^{6}\\ v=\frac{1}{7}x^{7} \]

\[ \frac{x^{7}ln(9x)}{7}-\int \frac{1}{7}x^{7}\frac{1}{x}dx\\ = \frac{x^{7}ln(9x)}{7}-\frac{1}{7} \int \frac{x^{7}}{x}dx\\ = \frac{x^{7}ln(9x)}{7}-\frac{1}{7}\int x^{6}dx\\ = \frac{x^{7}ln(9x)}{7}-\frac{1}{7}(\frac{x^{7}}{7})+C\\ = \frac{x^{7}ln(9x)}{7}-\frac{x^{7}}{49}+C \]

Answer: \(\frac{x^{7}ln(9x)}{7}-\frac{x^{7}}{49}+C\)

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.

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

\[ \begin{split} \int_1^{e^6}\frac{1}{6x} dx &= \frac{1}{6} ln(x)|_1^{e^6} \\ &= \frac{1}{6} ln(e^6) - \frac{1}{6} ln(1) \\ &= \frac{1}{6} \times 6 - \frac{1}{6} \times 0 \\ &= 1 \end{split} \]

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