Use integration by substitution to solve the integral below.
\(\int{4e^{-7x}dx}\)
Let \(u=-7x\), then \(du = -7dx\).
\[\begin{split} \int{4e^{-7x}dx} &= \int{\frac{-7 \times 4}{-7}e^{-7x}dx} \\ &= \int{\frac{-4}{7}e^u du} \\ &= \frac{-4}{7}e^u+constant \\ &= -\frac{4}{7}e^{-7x}+ constant \end{split}\]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.
\(N(1)= 6530\)
\[\begin{split} N(t) &= \frac{1050}{t^3}-220t+C \\ N(1) &= 6530 \\ \frac{3050}{1^3}-220\times 1 +C &= 6530 \\ C &= 6530 - 1050 + 220 \\ C &= 5700 \end{split}\]The level of contamination :
\(N(t) = \frac{1050}{t^3}-220t+5700\)
Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x)=2x-9\).
f = function(x){
(2*x) - 9
}
integrate(f,4.5,8.5)## 16 with absolute error < 1.8e-13Find the area of the region bounded by the graphs of the given equations.
\(y = x^2 - 2x - 2\)
\(y = x + 2\)
library(cubature)## Warning: package 'cubature' was built under R version 3.5.1f1 = function(x){
x+2
}
f2 = function(x){
(x^2)- (2*x) -2
}
hcubature(f1,-1,4)$integral - hcubature(f2,-1,4)$integral## [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.
Number of orders per year = n lot size = s Cost = c
ns = 110 s = 110 / n
c = 8.25n + 375/2n
c = 8.25n + 206.25/n
\(c'\) = 8.2 - 206.25/\(n^2\)
\(c'\) = 0
n = sqrt(206.25/8.25)
n = 5
5 orders per year.
Use integration by parts to solve the integral below.
\(\int{ln(9x) \times x^6 dx}\)
Let \(u= ln(9x)\)
\(\frac{du}{dx}=\frac{1}{x}\)
Let \(\frac{dv}{dx}=x^6\)
\(v = \int{x^6 dx} = \frac{1}{7}x^7\)
\(\int{u \frac{dv}{dx}dx} = uv - \int{v \frac{du}{dx} dx}\)
\[\begin{split} \int{ln(9x) \times x^6 dx} &= \frac{1}{7}x^7 \times ln(9x) - \int{\frac{1}{7}x^7 \times \frac{1}{x} dx} \\ &=\frac{1}{7}x^7 \times ln(9x) - \int{\frac{1}{7}x^6 dx} \\ &=\frac{7}{49}x^7 \times ln(9x) - \frac{1}{49}x^7 + constant \\ &=\frac{1}{49}x^7 (7ln(9x) - 1) + constant \\ \end{split}\]Determine whether \(f(x)\) is a probability density function on the interval \([1, e^6]\). If not, determine the value of the definite integral.
\(f(x) = \frac{1}{6x}\)
\[\begin{split} \int_1^{e^6}\frac{1}{6x} dx &= \frac{1}{6} ln(x)|_1^{e^6} \\ &= \frac{1}{6} ln(e^6) - \frac{1}{6} ln(1) \\ &= \frac{1}{6} \times 6 - \frac{1}{6} \times 0 \\ &= 1 \end{split}\]