1. Use integration by substitution to solve the integral below.

\(\int{4e^{-7x}dx}\)

solution let U = −7x

\(dU = −7dx\)

\(dx = \frac{dU}{−7}\)

\(4\int{e^U\frac{dU}{−7}}\)

\(-\frac{4}{7}∫e^UdU\)

\(\frac{-4}{7}e^U+C\)

\(-\frac{4}{7}e^{−7x}+C\)

2.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

solution

If we set N(1) = 6530, We can find The constant of integration, N0.

\(\frac{dN}{dt}=-\frac{3150}{t^4} - 220\)

\(dN =(\frac{3150}{t^4}−220)dt\)

\(N=\int{\frac{3150}{t^4}dt} −\int{220dt}\)

\(N = N_0 − \frac{3150}{3t^3}− 220t\)

\(N(_1) = N_0 −\frac{1050}{1^3}−220(1)\)

\(N_0 = 6530 + 1050 + 220\)

\(N_0 = 7800\)

\(N = 7800−\frac{1050}{t^3}−220t\)

3. Find the total area of the red rectangles in the figure below, where the 3. equation of the line is f ( x ) = 2x - 9.

solution

Using a R function to find the difference between the areas under the curve
fn = function(x) {2*x -9}
a <- integrate(fn, 4.5, 8.5)$value
(a <- round(as.numeric(a)))
## [1] 16

Find the area of the region bounded by the graphs of the given equations.

\(y = x2 - 2x - 2, y = x + 2\)

solution

using R functions to find the difference between areas under the curve
X = function(x) {x + 2}
Y = function(x) {x^2 -2*x -2}
ax <- integrate(X, -1, 4)
yx <- integrate(Y, -1, 4)
(Area <- round((ax$value - yx$value),4))
## [1] 20.8333

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

\(f′(x) = 1.875−\frac{907.5}{x^2} = 0\)

\(1.875 − \frac{907.5}{x^2} = 0\)

\(1.875 = \frac{907.5}{x^2}\)

\(1.875* x^2 = 907.5\)

\(x^2 = \frac{907.5}{1.875}\)

\(x = \sqrt{\frac{907.5}{1.875}}\)

\(x = 22\)

Use integration by parts to solve the integral below

\(\int{ln(9x).x^6 dx}\)

solution

\(=\frac{1}{7}x^7 * ln(9x) − \int{\frac{1}{7}x^7 * \frac{1}{x}dx}\)

\(=\frac{1}{7}x^7 * ln(9x) - \int{\frac{1}{7}x^6dx}\)

\(= \frac{7}{49}x^7 * in(9x) - \frac{1}{49} x^7dx + C\)

\(= \frac{1}{49}x^7*(7ln(9x) - 1) + C\)

7. Determine whether f ( x ) is a probability density function on the interval 1, e6 . If not, determine the value of the definite integral.

\(f(x) = \frac{1}{6x}\)

solution

\(F(x)=\int_1^{e^6}{f(x)dx}=1\)

where \(f(x) = \frac{1}{6x}\)

=> \(F(x) = \int{_1^{e^6}\frac{1}{6x}dx}\)

\(= \frac{1}{6}\int_1^{e^6}{\frac{1}{x}dx}\)

\(= \frac{1}{6}ln(x)|_1^{e^6}\)

\(= \frac{1}{6}[ln(e^6) - ln(1)]\)

\(= \frac{1}{6}[6-0]\)

=> \(F(x) = 1\)

At interval \([1, e^6]\), the integral function F(x) resolves to 1. Therefore, F(x) is a probability density function at this interval.