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

\[\int 4e^{-7x}dx\]

Solution:

\[\begin{equation} \int4e^{-7x}dx = 4 \int e^{-7x}dx \\\\ = 4\frac{e^{-7x}}{-7} + C \\\\ = -\frac {4} {7} e^{-7x} + C \\\\ \end{equation}\]

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.

Solution:

\[\begin{equation} \int \frac{dN}{dt} dt = -\int (\frac {3150}{t^4} + 220) dt\\\\ N(t) = (-3150\frac{t^{-3}}{3}) - 220t + C \end{equation}\]

since N(1) = 6530, we plug in t = 1 we get:

\[ N(1) = (3150\cdot \frac {1} {3}) - 220 + C = 6530 \\\\ C = 6530 + 220 - \frac {3150}{3} = 5260 \] this implies: \[N(t) = 1050t^{-3} - 220 + 5260\]

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

graph \[ \int_{4.5}^{8.5} (2x-9)dx = x^2 - 9x \\\\ = (8.5)^2-8.5*9 - ((4.5)^2 - 4.5*9) \\\\ = 16 \]

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

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

func_1 = function(x) {
  x^2 - 2*x - 2
}

func_2 = function(x) {
  x + 2
}

x <- seq(-2, 5, 0.05)

(y1 <- func_1(x))
##   [1]  6.0000  5.7025  5.4100  5.1225  4.8400  4.5625  4.2900  4.0225  3.7600
##  [10]  3.5025  3.2500  3.0025  2.7600  2.5225  2.2900  2.0625  1.8400  1.6225
##  [19]  1.4100  1.2025  1.0000  0.8025  0.6100  0.4225  0.2400  0.0625 -0.1100
##  [28] -0.2775 -0.4400 -0.5975 -0.7500 -0.8975 -1.0400 -1.1775 -1.3100 -1.4375
##  [37] -1.5600 -1.6775 -1.7900 -1.8975 -2.0000 -2.0975 -2.1900 -2.2775 -2.3600
##  [46] -2.4375 -2.5100 -2.5775 -2.6400 -2.6975 -2.7500 -2.7975 -2.8400 -2.8775
##  [55] -2.9100 -2.9375 -2.9600 -2.9775 -2.9900 -2.9975 -3.0000 -2.9975 -2.9900
##  [64] -2.9775 -2.9600 -2.9375 -2.9100 -2.8775 -2.8400 -2.7975 -2.7500 -2.6975
##  [73] -2.6400 -2.5775 -2.5100 -2.4375 -2.3600 -2.2775 -2.1900 -2.0975 -2.0000
##  [82] -1.8975 -1.7900 -1.6775 -1.5600 -1.4375 -1.3100 -1.1775 -1.0400 -0.8975
##  [91] -0.7500 -0.5975 -0.4400 -0.2775 -0.1100  0.0625  0.2400  0.4225  0.6100
## [100]  0.8025  1.0000  1.2025  1.4100  1.6225  1.8400  2.0625  2.2900  2.5225
## [109]  2.7600  3.0025  3.2500  3.5025  3.7600  4.0225  4.2900  4.5625  4.8400
## [118]  5.1225  5.4100  5.7025  6.0000  6.3025  6.6100  6.9225  7.2400  7.5625
## [127]  7.8900  8.2225  8.5600  8.9025  9.2500  9.6025  9.9600 10.3225 10.6900
## [136] 11.0625 11.4400 11.8225 12.2100 12.6025 13.0000
(y2 <- func_2(x))
##   [1] 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70
##  [16] 0.75 0.80 0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45
##  [31] 1.50 1.55 1.60 1.65 1.70 1.75 1.80 1.85 1.90 1.95 2.00 2.05 2.10 2.15 2.20
##  [46] 2.25 2.30 2.35 2.40 2.45 2.50 2.55 2.60 2.65 2.70 2.75 2.80 2.85 2.90 2.95
##  [61] 3.00 3.05 3.10 3.15 3.20 3.25 3.30 3.35 3.40 3.45 3.50 3.55 3.60 3.65 3.70
##  [76] 3.75 3.80 3.85 3.90 3.95 4.00 4.05 4.10 4.15 4.20 4.25 4.30 4.35 4.40 4.45
##  [91] 4.50 4.55 4.60 4.65 4.70 4.75 4.80 4.85 4.90 4.95 5.00 5.05 5.10 5.15 5.20
## [106] 5.25 5.30 5.35 5.40 5.45 5.50 5.55 5.60 5.65 5.70 5.75 5.80 5.85 5.90 5.95
## [121] 6.00 6.05 6.10 6.15 6.20 6.25 6.30 6.35 6.40 6.45 6.50 6.55 6.60 6.65 6.70
## [136] 6.75 6.80 6.85 6.90 6.95 7.00
plot(x, y1, type = 'l') +
  lines(x,y2, type = 'l')

## integer(0)

To determine the intersection points of the two curves, we set the two functions equal to each other and solve the resulting quadratic equation

\[ x^2 - 2x - 2 = x +2 \\\\ x^2 - 2x - 2 - x - 2 = 0 \\\\ x^2 - 3x - 4 = 0 \\\\ (x-4)(x+1) = 0 \] The intercepts occur at x=4 and x=−1 which allows us to integrate the difference of the two functions to get the area of the region.

The area bounded by the two curve is the integral:

\[ \int_{-1}^4 (x+2) - (x^2 - 2x - 2)dx = \int_{-1}^4 -x^2 +3x + 4 \space dx \\\\ = (-\frac {x^3}{3} + \frac{3}{2}x^2 + 4x)|_{-1}^{4} \\\\ = (-\frac{4^3}{3} + \frac{3}{4^2} + 4(4) - (-\frac{(-1)^3}{3} + \frac {3}{2}\cdot(-1)^2 + 4(-1)) \\\\ = 20\frac{5}{6} \\\\ = 20.8333 \]

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.

solution:

f(x)=8.25x+(110/x)3.75, where x is number of orders, 0<x<111

f’(x) = 8.25−(110/x^2)3.75=0

\(x = \sqrt{110∗3.75/8.75}\) = 6.9

(110*3.75/8.75)^.5
## [1] 6.866066
test <- c()

for (i in (1:110))
{test<-c(test,(8.25*i+110*3.75/i))
}

match(min(test),test) 
## [1] 7
6. Use integration by parts to solve the integral below.

\(\int ln(9x)\cdot x^6 \space dx\)

\(Sol^n\)

\[ u=x7; \space u′=7x6; \space v=ln(9x)/7 \\\\\ \int ln(9x)\cdot x^6 \space dx = \int vdu = vu - \int udv \\\\ = x^7ln(9x)/7 - \int x^7dln(9x)/7 \\\\ = x^7ln(9x)/7 - \int x^6/7\space dx \\\\ = x^7 ln(9x)/7 - x^7/49 + c \\\\ = x^7ln(9x)/7 - x^7/49 + c \]

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.

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

\(Sol^n\)

\[ \int_1^6 \frac{1}{6}xdx = ln\frac{x}{6} = \frac{6}{6} = 1. \]

as integral of our function is equal to 1 on interval \([1,e^6]\), f(x) is a probability density function.