Akula-DATA605-Week13-HW13
Pavan Akula
November 25, 2017
Answer:
Using substitution method
\(\int 4e^{-7x}dx\)
Let \(u = -7x\)
\(du = -7dx\)
\(-\frac{du}{7} = dx\)
Applying to actual equation,
\(\int 4e^{u} \times -\frac{du}{7}\)
\(\int -\frac{4}{7}e^u du\)
\(-\frac{4}{7} \int e^u du\)
\(-\frac{4}{7} e^u + C\)
Substituting \(u\)
\(-\frac{4}{7} e^{-7x} + C\)
Answer:
Change of bacteria rate
\(\frac{dN}{dt} = -\frac{3150}{t^4}-220\)
Solving for \(N\)
\(dN = (-\frac{3150}{t^4}-220)dt\)
\(\int dN = \int(-\frac{3150}{t^4}-220)dt\)
Using substitution method
Let \(u = \frac{1}{t}\)
\(\int u^4du = \frac{u^{n+1}}{n+1}\)
Applying the values
\(N = \int(-\frac{3150}{t^4}-220)dt\)
\(N = \int-3150{t^{-4}}dt-\int220dt\)
\(N = -3150(\frac{t^{-4+1}}{-4+1}) - 220t\)
\(N = -3150(\frac{t^{-3}}{3}) - 220t\)
\(N = -3150(\frac{1}{3t^3}) - 220t\)
\(N = -\frac{3150}{3t^3} - 220t\)
\(N = -\frac{1050}{t^3} - 220t\)
Level of contamination on \(1^{st}\) day, \(t = 1\)
\(N_1 = \frac{-3150}{3\times1^3} - 220\times 1\)
\(N_1 = -1050 - 220\)
\(N_1 = -1270\)
Level of contamination on \(n^{th}\) day is \(N(t) = Level\ before\ treatment - N_1 - N\)
\(N(t) = 6530 - N_1 - N\)
\(N(t) = 6530 -( -1270) - (-\frac{1050}{t^3} - 220t)\)
\(N(t) = 7800 +\frac{1050}{t^3} + 220t\)
Answer:
From the diagram, \(x\) varies from \(4.5\) to \(8.5\)
Given equation \(f(x) = 2x - 9\),
Since \(x\) is changing
\(\int(fx) dx = \int_{4.5}^{8.5} 2x-9\)
= \(2\times\frac{x^2}{2} - 9x|_{4.5}^{8.5}\)
= \(x^2 - 9x|_{4.5}^{8.5}\)
= \((8.5^2 - 9\times8.5) - (4.5^2 - 9\times4.5)\)
= \(16\)
Area of the rectangles = \(16\) units.
Answer:
Given equations,
\(y = x^2-2x-2\)
\(y = x+2\)
Since \(y\) is defined by two different equations, lets solve \(x\),
\(x^2-2x-2 = x+2\)
= \(x^2 -3x - 4 = 0\)
Using polynomial division ,
\((x + 1)(x-4) = 0\)
Hence \(x = {-1, 4}\)
Area between the curves
\(\int^4_{-1}(x^2-2x-2)dx - \int^4_{-1}(x+2)dx\)
\(\int^4_{-1}(x^2-2x-2 - (x+2))dx\)
\(\int^4_{-1}(x^2-2x-2 - (x+2))dx\)
\(\int^4_{-1}(x^2-3x-4)dx\)
\(\bigg[\frac{x^3}{3}-\frac{3x^2}{2}-4x\bigg]_{-1}^{4}\)
\(\bigg[\frac{4^3}{3} - \frac{3\times4^2}{2} - 4\times4\bigg] - \bigg[\frac{(-1)^3}{3} - \frac{3\times(-1)^2}{2} - 4\times(-1)\bigg]\)
\(\bigg[\frac{64}{3} - \frac{48}{2} - 16\bigg] - \bigg[\frac{-1}{3} - \frac{3}{2} + 4\bigg]\)
x = 4
eq1 = ((x^3)/3) - (3*(x^2)/2) - (4*x)
x = -1
eq2 = ((x^3)/3) - (3*(x^2)/2) - (4*x)
#area cannot be negative
area = abs(eq1 - eq2)Area between curves = \(20.8333333\) units
Answer:
There are two ways to look at the problem
- To sell the product fast, such that average time flat iron is stored is less than a year (let’s say reduce to 6 months) resulting in lower storage costs.
- If flat irons are selling fast increase order size, to lower fixed costs.
In both cases selling fast is key. Hence we have to fine-tune storage costs.
Expected flat iron sales = \(110\) units
Let o be required orders and x be number of units per order. Such that \(o \times x = 110\), \(x = \frac{110}{o}\)
Order cost = \(8.25 \times o\), Storage costs = \(3.75 \times x\)
Total cost(\(C\)) = \(8.25 \times o + 3.75 \times x\)
Since we want to reduce storage costs in half
Total cost(\(C\)) = \(8.25 \times o + \frac{3.75 \times x}{2}\)
= \(8.25o + 1.875 x\)
= \(8.25o + 1.875 \times \frac{110}{o}\)
= \(8.25o + \frac{206.25}{o}\)
Using first derivative\(C^{'} = 0\),
= \(\frac{d}{do}(8.25o + \frac{206.25}{o})\)
= \(8.25 - \frac{206.25}{o^2}\)
\(o^2 = \frac{206.25}{8.25}\)
\(o^2 = 25\)
\(o = \sqrt25\)
\(o = 5\)
So number of orders store has to place to reduce storage costs in half = \(5\)
Answer:
\(\int ln(9x).x^6dx\)
Using product of function and derivative of another function
\(\int f(x).g^{'}(x)dx = f(x).g(x) - \int f^{'}(x)g(x)dx\)
\(f(x) = ln(9x)\), \(g^{'}(x) = x^6\)
derivative of \(f(x)\)
\(f^{'}(x) = \frac{d}{dx}ln(9x) = \frac{1}{x}\)
\(g^{'}(x) = x^6\)
integral of \(g^{'}(x)\)
\(\int g^{'}(x) = g(x) = \int x^6 = \frac{x^7}{7}\)
\(f(x).g(x) - \int f^{'}(x)g(x)dx\)
= \(ln(9x)\frac{x^7}{7} - \int \frac{1}{x}\times\frac{x^7}{7}\)
= \(ln(9x)\frac{x^7}{7} - \int \frac{x^6}{7}\)
= \(ln(9x)\frac{x^7}{7} - \frac{1}{7}\int x^6\)
= \(ln(9x)\frac{x^7}{7} - \frac{1}{7}\times \frac{x^7}{7}\)
= \(\frac{x^7}{7} (ln(9x)- \frac{1}{7})\)
Answer:
\(\int_1^{e^6}\frac{1}{6x}dx\)
= \(\frac{1}{6}\int_1^{e^6}\frac{1}{x}dx\)
= \(\frac{1}{6}\int_1^{e^6}\frac{1}{x}dx\)
= \(\frac{1}{6}\bigg[ln(x)\bigg]_1^{e^6}\)
= \(\frac{1}{6}\bigg[ln(e^6) - ln(1)\bigg]\)
= \(\frac{1}{6}\bigg[6 - 0\bigg]\)
= \(1\)