Use integration by substitution to solve the integral below
\(\int_{}^{} 4e^{-7x} \, dx\)
let \(u=-7x\)
\(du=-7 dx\)
\(dx=-\frac{1}{7}du\)
Now substitute back into the integral:
\(\int_{}^{} 4e^{u} * -\frac{1}{7} \, du\)
\(-\frac{4}{7} \int_{}^{} e^{u} \, du=\)
\(=-\frac{4}{7}e^u + c\)
Substitute u = -7x
\(=-\frac{4}{7}e^{-7x} + c\)
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.
To find the function N(t) we need to integrate the given rate of change
\(N(t) = \int_{}^{} (-\frac{3150}{t^4} - 220) \, dt\)
\(N(t) = \int_{}^{} (-(\frac{1}{t^4} * 3150) - 220) \, dt\)
\(N(t) = (-\frac{1}{3t^3} * 3150) - 220t + C\)
\(N(t) = \frac{1050}{t^3} - 220t + C\)
Need to plug in N(1) = 6530
\(6530 = \frac{1050}{1^3} - 220(1) + C\)
\(C=5700\)
So N(t) becomes:
\(N(t) = \frac{1050}{t^3} - 220t + 5700\)
Find the total area of the red rectangles in the figure below, where the equation of the line is f(x) = 2x - 9
The picture in the assignment shows the rectnagles going from x= 4.5 to x=8.5
So to find the area under the line we will integrate the function from 4.5 to 8.5
\(\int_{4.5}^{8.5} 2x-9 \, dx\)
We can use r to calculate this integral
f <- function(x) { 2*x - 9 }
# Perform the integration over the interval from 4.5 to 8.5
result <- integrate(f, lower = 4.5, upper = 8.5)
# Display the value of the integration
print(result$value)## [1] 16The area of the rectangles is 16
Find the area of the region bounded by the graphs of the given equations.
\(y=x^2-2x-2, y=x+2\)
We need to find where the two functions intersect so set them equal to each other and then solve.
\(x^2-2x-2 = x+2\)
\(x^2-3x-4=0\)
\((x-4)(x+1)=0\)
\(x=4,x=-1\)
Now we know that the functions intersect at x=4 and x=-1 so we can subtract the two functions and integrate from x=-1 to x=4
\(\int_{4}^{-1} (x+2) - (x^2-2x-2) \, dx\)
\(\int_{4}^{-1} -x^2+3x+4 \, dx\)
Now we can use R to solve the integral
f <- function(x) { -x^2+3*x+4 }
result <- integrate(f, lower = -1, upper = 4)
print(result$value)## [1] 20.83333The area of the region bounded by the grpahs is 20.83333
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.
We want to try and find the optimal size and the optimal number of orders per year for the beauty supply company.
For the whole year since we know that they expect to sell 110 flat irons this means that:
number of orders * lot size = 110
Solving for lot size :
lot size = \(\frac{110}{number\ of\ orders}\)
\(f(x) = 8.25x + 3.75 * \frac{\frac{110}{x}}{2}\)
\(f(x) = 8.25x+ 3.75 * \frac{55}{x}\)
\(f(x) = 8.25x + \frac{206.25}{x}\)
Now differentiate the function and solve when it = 0
\(f'(x) = 8.25 -\frac{206.25}{x^2}\)
Now set it equal to 0 and solve for x
\(8.25 - \frac{206.25}{x^2} = 0\)
\(8.25 = \frac{206.25}{x^2}\)
\(x^2=\frac{206.25}{8.25}\)
\(x = 5\)
The number of orders per year that will minimize cost is 5 and therefore the lot size should be 110/5 = 22 flat irons per order
Use integration by parts to solve the integral below.
\(\int_{}^{} ln(9x)*x^6 \, dx\)
Integration by parts:
\(\int_{}^{} u\, dv= uv- \int_{}^{} v\, du\)
\(Choose\ u = ln(9x), \ dv= x^6 dx\)
\(du = \frac{1}{x}dx\)
\(v = \frac{x^7}{7}\)
\(\int_{}^{} ln(9x) * x^6 \,dx = ln(9x) * \frac{x^7}{7} - \int \frac{x^7}{7} *\frac{1}{x} \,dx\)
\(\int ln(9x) * x^6 \,dx = \frac{x^7 ln(9x)}{7} - \frac{1}{7} \int x^6 \,dx\)
\(\frac{x^7 ln(9x)}{7} - \frac{x^7}{49} + C\)
Determine whether f(x) is a probability density function on the interval [1, e6] . If not, determine the value of the definite integral.
\(f(x) = \frac{1}{6x}\)
If f(x) is a probability density function on the interval given then the integral on that interval will equal 1 so we can use R to compute the integral and if it equals 1 we know its a PDF.
f <- function(x) { (1/(6*x)) }
result <- integrate(f, lower = 1, upper = exp(6))
print(result$value)## [1] 1Using R we can see that the integral evaluates to 1 therefore this function is a PDF on the interval \([1,e^6]\)