## Warning: package 'Deriv' was built under R version 3.3.3
## Warning: package 'rJava' was built under R version 3.3.3
## installed_and_loaded.packages.
## prettydoc TRUE
## Deriv TRUE
## rootSolve TRUE
## rSymPy TRUE1. Use integration by substitution to solve the integral below.
\(\int 4e^{-7x} dx\)
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.
ft <- function(t) -3150 * t^-4 - 220
# anti-derivative
# F(t) = 1050 * t^-3 - 220 * t + C
# solve for C:
# F(1) = 6530 = 1050 * t^-3 - 220 * t + C
t <- 1
(C <- 6530 - 1050 * t^-3 + 220 * t)## [1] 5700# Integral
(Ft <- function(t) 1050 * t^-3 - 220 * t + 5700)## function(t) 1050 * t^-3 - 220 * t + 57003. Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x) = 2x-9\)
fx <- function(x) 2 * x - 9
# f(0) = 2x-9
# x = 4.5
(lower <- 4.5)## [1] 4.5(upper <- lower + 4)## [1] 8.5stats::integrate(Vectorize(fx), lower, upper)## 16 with absolute error < 1.8e-134. Find the area of the region bounded by the graphs of the given equations.
\(y = x^2 -2x -2\)
\(y = x + 2\)
s <- seq(-2, 5, by = 0.01)
fx <- function(x) x^2 - 2 * x - 2
fx2 <- function(x) x + 2
fx_values <- fx(s)
fx2_values <- fx2(s)
plot(s, fx_values, type = "l", col = "red")
lines(s, fx2_values, col = "green", type = "l")
abline(0, 0)
# intersections
is_equal <- fx2_values == fx_values
mydf <- data.frame(cbind(s, is_equal))
intersections <- subset(mydf, is_equal == 1)
# critical points
roots <- uniroot.all(fx, c(-2, 5))
lower1 <- roots[1]
upper1 <- roots[2]
lower2 <- intersections$s[1]
upper2 <- intersections$s[2]
# answer
(integrate(Vectorize(fx2), lower2, upper2)$value - integrate(Vectorize(fx),
lower2, upper2)$value)## [1] 20.83333### Validate answer###
Fx_neg <- abs(integrate(Vectorize(fx), lower1, upper1)$value)
Fx_pos1 <- integrate(Vectorize(fx), lower2, lower1)$value
Fx_pos2 <- integrate(Vectorize(fx), upper1, upper2)$value
Fx2 <- integrate(Vectorize(fx2), lower2, upper2)$value
Fx_neg + Fx2 - Fx_pos2 - Fx_pos1## [1] 20.833335. 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.
# lotsize = 110 / orders_per_year
# cost = lotsize*3.75 + orders_per_year * 8.25
total_cost <- function(orders_per_year) 110/orders_per_year * 3.75 + 8.25 *
orders_per_year
(total_cost.prime <- Deriv(total_cost))## function (orders_per_year)
## 8.25 - 412.5/orders_per_year^2(root <- uniroot.all(total_cost.prime, c(1, 110)))## [1] 7.071079# calculate in whole number terms
orders_per_year <- floor(root)
lot_size <- ceiling(110/orders_per_year)
total_units <- orders_per_year * lot_size
min_cost <- lot_size * 3.75 + 8.25 * orders_per_year
paste(orders_per_year, "orders of", lot_size, "units each,", total_units, "total units, for a minimum cost of",
min_cost)## [1] "7 orders of 16 units each, 112 total units, for a minimum cost of 117.75"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, e^6] . If not, determine the value of the definite integral.
\(f ( x ) = 1/6x\)
fx <- function(x) 1/(6 * x)
lower <- 1
upper <- exp(6)
all.equal(1, integrate(Vectorize(fx), lower, upper)$value, tolerance = 10^-8)## [1] TRUE