1. Use integration by substitution to solve the integral below.∫4e−7xdx
Solution :
let u = -7x
du/dx = -7
dx = du/-7
By substitution -
∫4e−7xdx
= 4∫eudu/−7 = −4/7∫eudu = -4/7 * eu + C
= -4/7 * e−7x + C
2. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of dN/dt = -3150 / t4 - 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:
First find the general function for N(t) -
∫−3150/t4−220dt = -3150 * (t−3/-3) - 220t + C = N
After 1 day, N = 6530, solving for C
(C <- 6530 + 3150 * (1^-3)/-3 + 220*(1))
## [1] 5700
N(t) = -3150 * (t−3/-3) - 220t + 5700 = 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.
Solution :Each square in the graph has an area of 1. Each rectangle has a width of 1. Counting each rectangle left to right the areas are :
(Area=1+3+5+7)
## [1] 16
4. Find the area of the region bounded by the graphs of the given equations. y = x2 - 2x - 2, y = x + 2
#Find area in-build function
f1 = function(x) {x + 2}
f2 = function(x) {x^2 -2*x -2}
#Find the difference between areas under the curve
area1 <- integrate(f1, -1, 4)
area2 <- integrate(f2, -1, 4)
(area <- round((area1$value - area2$value),4))
## [1] 20.8333
# plot
curve(x^2 - 2*x - 2, -5, 5, col = "green")
curve(x + 2, -5, 5, add=T, col="orange")
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 :
Let x be a number of flat irons,
Yearly storage cost = Storage cost per iron × Average number of irons =3.75.x2=1.875x
Yearly ordering cost = Cost of each order × Number of orders = 8.25.110x=907.5x
lot.size=x=22,order=11022=5
The lot size is 22 and the number of orders per year of 5 will minimize inventory costs.
6. Use integration by parts to solve the integral below. ∫ln(9x)x6dx
Solution :
∫ln(9x)×x6dx
=1/7x7×ln(9x)−∫1/7x7×1xdx
=1/7x7×ln(9x)−∫1/7x6dx
=7/49x7×ln(9x)−1/49x7+c
=1/49x7(7ln(9x)−1)+c
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) = 1/6x
Solution:
∫e611/(6x)dx
1/6 * ln x from [1, e6], =
(PDF <- (1/6) * log(exp(6)) - ((1/6) * log(1)))
## [1] 1
f(x) is a probability density function on the interval [1, e6] - with the area of the curve as 1 (100% probabily for the interval)