Use integration by substitution to solve the integral below.
\[ \int_{}^{}4e^{-7x}dx \] Substitute
\[ u= −7x \to \frac{du}{dx} = -7 \to dx = -\frac{1}{7}du \]
\[ -\frac{4}{7}\int_{}^{}e^{u}du = -\frac{4}{7}e^u \] Substitute \[ u= −7x \to -\frac{4}{7}e^{-7x} \] \[ \int_{}^{}4e^{-7x}dx = -\frac{4}{7}e^{-7x} + C \]
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
Integrate the rate of change of contamination with respect to time t to find N(t).
\[ \frac{dN}{dt} = -\frac{3150}{t^4} - 220 \] \[ dt\frac{dN}{dt} = (-\frac{3150}{t^4} - 220) dt \] \[ dN = (-\frac{3150}{t^4} - 220) dt \to \int_{}^{}dN dt = \int_{}^{}-\frac{3150}{t^4} - 220 dt \]
\[ N(t) = \frac{-3150}{-3t^3} - 220t \] \[ N(t) = \frac{1050}{t^3} - 220t + C \] After 1 day was 6530 bacteria per cubic centimeter: \[ 6530 = \frac{1050}{(1)^3} - 220(1) + C \]
Solve for C: \[ C = 6530 + 220 - 1050 = 5700 \]
The function would therefore be: \[ N(t) = \frac{1050}{t^3} - 220t + 5700 \] ### Question 3
Find the total area of the red rectangles in the figure below, where the equation of the line is f (x) = 2x - 9.
f <- function(x) {2*x-9}
f_area <- integrate(f,4.5,8.5)
f_area$value## [1] 16Find the area of the region bounded by the graphs of the given equations \(y = x^2 -2x -2\) , \(y = x + 2\)
f <- function(x) {x^2 -2*x - 2}
g <- function(x) {x + 2}
#retrieve the points where the functions intersect below zero
rt1 <- uniroot(function(x) f(x) - g(x) , c(-5,0), tol=1e-8)
rt1$root## [1] -1#retrieve the points where the functions intersect above zero
rt2 <- uniroot(function(x) f(x) - g(x) , c(0,5), tol=1e-8)
rt2$root## [1] 4#calculate the area under the f(x)
fx_area <- integrate(f, lower = -1, upper = 4)
#calculate the area under the g(x)
gx_area <- integrate(g, lower = -1, upper = 4)
#Find the difference
(gx_area$value - fx_area$value)## [1] 20.83333A 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.
Inventory cost
= Storage Cost + Order Cost
Storage cost assuming half of inventory
is in storage:
\[ 3.75 (\frac{x}{2}) \]
Order cost: \[ 8.25 (\frac{110}{x}) \] Inventory cost: \[ f(x) = 3.75 (\frac{x}{2}) + 8.25 (\frac{110}{x}) \] To minimize we take the derivative of the function and set the inventory cost to zero and solve for x. \[ f(x)' = 3.75 (\frac{1}{2}) - 8.25 (\frac{110}{x^2}) \]
a = 3.75/2
b = 8.25*110
(x = sqrt(b/a))## [1] 22To solve for the number of orders per year to minimize inventory
(110/x)## [1] 5Use integration by parts to solve the integral below.
\[ \int_{}^{}\ln(9x)\cdot x^6 dx \] Integration by Parts: \[ \int_{}^{}fg = fg' - \int_{}^{}f'g \]
\[ f = \ln(9x) \]
\[ f' = \frac{1}{x} \]
\[ g = x^6 \]
\[ g' = \frac{x^7}{7} \]
\[ \frac{\ln(9x)x^7}{7} - \int_{}^{}\frac{x^6}{7}dx \]
\[ \frac{\ln(9x)x^7}{7} - \frac{x^7}{49} + C \]
Determine whether f ( x ) is a probability density function on the interval \([1, e^6]\). If not, determine the value of the definite integral. \[ f(x) = \frac{1}{6x} \]
For a function to be a probability density function it must satisfy
the following conditions:
The probability density function is
non-negative for all the possible values
The area between the
density curve and horizontal X-axis is equal to 1
f <- function(x){1 / (6 * x)}
f_area = integrate(f, lower=1, upper=exp(6))
f_area$value## [1] 1The function satisfy the probability density function conditions.