Univariate & Multivariate Calculus

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

\(\int4e^{-7x}dx\)

Solution:

Key:

u = -7x

du = -7dx

\(\int4e^{-7x}dx = \int \frac{-7x4}{-7} e^{-7x} dx\)

\(=\int\frac{-4}{7} e^{u} du\)

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

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

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2. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \(\frac{dN}{dt} = \frac{3150}{t^4}\) 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.

Disclaimer: I am not 100% on the approach or solution but attempted to understand it and solve it.
Given:

\(\frac{dN}{dt} = N'(t)\)

Solution:

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

\(N(t) = \int(\frac{-3150}{t^4} -220)dt=\frac{1050}{t^3}-200t + C\)

\(N(1) = 6530\)

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

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

\(C = 6530 - 1050 +220\)

C = 5700

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

Solution:

f(x) = 2x - 9

\(\int_{4.5}^{8.5} 2x = 9dx\)

\(=[x^2 - 9x]|^{8.5}_{4.5}\)

\(=(8.5^2 - 9(8.5))-(4.5^2 - 9(4.5))\)

Area of red rectangles = 16

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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

# Create funtions
y1 <- function(x){x^2 - (2*x) - 2}
y2 <- function(x){x + 2}

# plot lines
curve(y1, from = -6, to = 6, col = 2)
curve(y2, from = -6, to = 6, add = TRUE, col = 4)

integrate(y1, lower = -1, upper = 4)
## -3.333333 with absolute error < 1.2e-13
integrate(y2, lower = -1, upper = 4)
## 17.5 with absolute error < 1.9e-13
total_area = 17.5 - (-3.33)
total_area
## [1] 20.83

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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.

Solution:

n = number orders per year

x = number of flat irons to order per year

x / 2 = average inventory level

Storage cost = 3.75 * (x / 2) Storage cost = 1.875x

Order cost = 8.25 * 110 / x Order cost = 907.5 / x

Inventory cost = 1.875x + (907.5 /x)

Therefore:

1.875 - 907.5 * x - 2 = 0

x = \(\sqrt484\)

x = 22

Orders per year = 110 / n = 22

n = 5

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6. Use integration by parts to solve the integral below

\(\int ln(9x) * x^6 dx\)

Solution:

u = In(9x) therefore du = \(\frac{1}{x}dx\)

v = \(\frac{1}{7}x^7\) therefore \(dv = x^6dx\)

With integration we get \(\int uv' = uv = \int u'v\)

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

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

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

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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.

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

Solution:

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

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

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