MSDS Spring 2018

DATA 605 Fundamentals of Computational Mathematics

Jiadi Li

HW #13 - Applications of Derivative and Integration

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

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

Let \(u\) = \(-7x\), then \(du\) = \(-7dx\) and \(dx\) = \(-\frac{1}{7}du\)

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

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.

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

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

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

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

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

\(\because N(1)=6530\)

\(\therefore N(1) = -\frac{1050}{1^3}-220(1)+C = 6530\space\space\) \(\therefore \space C = 6530 + 1050 + 220 = 7800\)

\(\Rightarrow N(t) = -\frac{1050}{t^3}-220t+7800\)

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

The total area of the red rectangles includes the area of 4 rectangles:
\(A = \int_{4.5}^{8.5}(2x-9)dx = x^2 + 9x |_{4.5}^{8.5} = 16\)

fx <- function(x){2*x-9}
integrate(fx,4.5,8.5)

## 16 with absolute error < 1.8e-13

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

find intersections: \(x^2-2x-2\) = \(x+2\) \(\Rightarrow\) \(x^2-3x-4\) = 0 \(\Rightarrow\) \((x-4)(x+1)\) = 0
\(\Rightarrow\) \(x_1\) = -1, \(x_2\) = 4

\(A = \int_{-1}^{4}(x+2) - (x^2 -2x -2)dx = -\frac{x^3}{3}+\frac{3x^2}{2}+4x|_{-1}^{4} = \frac{125}{6}\)

fx <- function(x){(x+2) - (x^2-2*x-2)}
integrate(fx,-1,4)

## 20.83333 with absolute error < 2.3e-13

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.

lot size \(\times\) number of orders per year = 110 flat irons
total cost = $3.75 \(\times\) lot size + $8.25 \(\times\) number of orders per year = $3.75 \(\times\) lot size + $8.25 \(\times \frac{110}{lot\space size}\)

\(C(l) = 3.75l + \frac{907.5}{l}\)

\(C'(l) = 3.75 - \frac{907.5}{l^2} = 0\)
\(l = \sqrt{\frac{907.5}{3.75}} \approx 16\) \(Rightarrrow\) \(\frac{110}{l} \approx 7\)

6.Use integration by parts to solve the integral below.
\(\int{ln(9x)\cdot x^6dx}\)

integration by parts: \(\int udv = uv-\int vdu\)

Let \(u = ln(9x)\), then \(du = \frac{1}{x}dx\)

Let \(dv = x^6dx\), then \(v = \frac{x^7}{7}\)

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

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

probability density function: \(F(x) = \int_{a}^{b}f(x)dx = 1\)

\(\because F(x) = \int_{1}^{e^6}\frac{1}{6x}dx = \frac{1}{6}ln(x)|_{1}^{e^6} = \frac{1}{6}[ln(e^6) - ln(1)] = \frac{1}{6}[6 - 0] = 1\)

\(\therefore f(x)\) is a probability density function on the interval [1,\(e^6\)].