1. Use integration by substitution to solve the integral below

\(∫4e^{−7x}dx\)

Answer

Let u = -7x

\(\frac{du}{dx} = \frac{d(-7x)}{dx} = -7\)

\(du = -7dx\)

\(dx = -\frac{1}{7}du\)

Replacing dx and u in \(∫4e^{−7x}dx\) gives us,

\(-\frac{1}{7}∫4e^{u}du = -\frac{4}{7}e^{-7x} + C\)

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.

Answer

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

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

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

\(= -3150∫t^{-4}dt - 220∫dt\)

\(N(t)=-3150\frac{-1}{3}t^{-3}-220t = \frac{1050}{t^3} - 220t + C\)

Given \(N(1) = 6530\), so for level after 1 day,

\(6530 = \frac{1050}{1^3} - 200(1) + C = 6530 = 1050 - 220 + C\)

\(C= 6530 - 1050 + 220 = 5700\)

Hence, \(N(t) = \frac{1050}{t^3} - 220t + 5700\)

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

problem3.

Answer

\(x = 4.5 to 8.5\)

#Find area in-build function
f3 = function(x) {2*x -9}

#Find the difference between areas under the curve
area3 <- integrate(f3, 4.5, 8.5)$value
area3 <- round(as.numeric(area3))
print(area3)
## [1] 16

4. Find the area of the region bounded by the graphs of the given equations.\(y = x^2 - 2x - 2, y = x + 2\) Enter your answer below.

Answer

#Find area in-build function
f1 = function(x) {x + 2}
f2 = function(x) {x^2 -2*x -2}

#Find the difference between areas under the curve
area1 <- integrate(f1, -1, 4)
area2 <- integrate(f2, -1, 4)
area <- round((area1$value - area2$value),4)
print(area)
## [1] 20.8333

5. 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.

Answer

\(f′(x) = 1.875 − \frac{907.5}{x^2}\)

\(f′(x) = 0\)

\(1.875 − \frac{907.5}{x^2} = 0\)

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

\(1.875x^2 = 907.5\)

\(x^2 = \frac{907.5}{1.875} = \sqrt\frac{907.5}{1.875} = \sqrt{484} = 22\)

The number of orders per year to minimize inventory costs is \(\frac{110}{n} = \frac{110}{22} = 5\)

6. Use integration by parts to solve the integral below.

\(∫ln(9x)x^6dx\)

Answer

Let \(u=ln(9x), du= \frac{1}{x}, v=\frac{1}{7}x^7, dv=x^6\)

\(∫udv = uv - ∫vdu\)

\(ln(9x)\frac{1}{7}x^7 - ∫\frac{1}{x}\frac{1}{7}x^7dx =\frac{1}{7}ln(9x)x^7 - \frac{1}{7}∫\frac{x^7}{x}dx\)

\(=\frac{1}{7}ln(9x)x^7 - \frac{1}{7}∫\frac{x^6}dx = \frac{1}{7}ln(9x)x^7 - \frac{1}{7}{1}{7}x^7 + C\) \(=\frac{1}{7}ln(9x)x^7 - \frac{1}{49}x^7 + C\)

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

Answer

\(∫_1^{e^6} \frac{1}{6x}dx = \frac{1}{6}ln(x)∫_1^{e^6}\) \(=\frac{1}{6}ln(e^6) - \frac{1}{6}ln(1) = \frac{1}{6}*6 - \frac{1}{6}*0 = 1\)

The total area under the interval \([1, e^6]\) is equal 1 hence f(x) is a probability density function.