Use integration by substitution to solve the integral below.
\[ \int 4e^{-7x}dx \]
\[ u = -7x \\ du = -7 dx \rightarrow \frac{-1}{7}du = dx\\ \int 4e^{-7x}dx \rightarrow \text{ Substitute } \rightarrow 4 \int e^u \frac{-1}{7} du\\ = \frac{-4}{7} \int e^u du = \frac{-4}{7} e^u + C \\ = \frac{-4}{7} e^{-7x} + C \]
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.
\[ \frac{dN}{dt} = -\frac{3150}{t^4} - 220\\ dN = \int \frac{3150}{t^4} dt - \int 220 dt\\ = \int 3150 t^{-4} dt - \int 220 dt\\ = \frac{3150}{t^-3} - 220t + C\\ N(t) = \frac{1050}{t^3} - 220t + C\\ N(1) = \frac{1050}{(1)^3} - 220(1) + C = 6350\\ 830 + C = 6350\\ C = 5520\\ N(t) = \frac{1050}{t^3} - 220t + 5520\\ \]
Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x)=2x-9\).
We need to integrate between the limits of a and b. a is 4.5 and b is 8.5.
\[ a=4.5 \text{ and } b=8.5\\ f(x) = 2x-9\\ \int_{4.5}^{8.5} 2x - 9 dx =\\ [x^2 - 9x]_{4.5}^{8.5} \\ = [8.5^2 - 9(8.5)] - [4.5^2 - 9(4.5)]\\ = -4.25 - (-20.25) \\ = 16\\ \]
Find the area of the region bounded by the graphs of the given equations. \[ y = x^2-2x-2\\ y = x+2\\ \] ### Solution
First we need to find the points of intersection:
\[ x^2-2x-2=x+2\\ x^2-3x-4=0\\ (x-4)(x+1)=0\\ x=4 \text{ and } x=-1\\ \] The points of intersection are \(x = -1\) and \(x=4\). \(x=-1\) is the lower limit (a) and \(x=4\) is the upper limit (b).
Now we will solve for the area between the two curves:
\[ A = \int_a^b [\text{Top } - \text{ Bottom}] dx\\ = \int_{-1}^{4} x+2 - (x^2 - 2x - 2) dx\\ = int_{-1}^{4} x+2 - x^2 + 2x + 2 dx\\ = int_{-1}^{4} -x^2 + 3x + 4 dx \\ = -\frac{x^3}{3} + \frac{3}{2} x^2 + 4x |_{-1}^{4}\\ = [-\frac{(4^3)}{3} + \frac{3}{2}(4)^2 + 4(4)] - [-\frac{(-1)^3}{3} + \frac{3}{2}(-1)^2 + 4(-1)] \\ = [\frac{-64}{3} +24 + 16] - [\frac{1}{3} +\frac{3}{2} -4]\\ = 20.8333 \]
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.
# demand
d = 110
# order cost
s = 8.25
# storage cost
h = 3.75
# lot size
q = sqrt((2*d*s)/h)
print(q)## [1] 22# number of orders per year
n = d/q
print(n)## [1] 5Use integration by parts to solve the integral below.
\(\int ln(9x) \cdot x^6 dx\)
\[ \int ln(9x) \cdot x^6 dx = \int [ln(9) + ln(x)] \cdot x^6 dx =\\ \int ln(9) x^6 dx + \int ln(x) x^6 dx\\ \text{Integrate the first integral: }\\ \frac{ln(9)}{7}x^7\\ \text{Integrate the second integral: }\\ u = ln(x)\\ du = \frac{1}{x} dx\\ dv = x^6 dx\\ v = \frac{x^7}{7}\\ \frac{x^7 ln(x)}{7} - \int \frac{x^7}{7} \frac{1}{x}dx=\\ \frac{x^7 ln(x)}{7} - \int \frac{x^6}{7} dx\\ = \frac{x^7 ln(x)}{7}- \frac{x^7}{49} + C\\ \text{Combine the two results: }\\ = \frac{x^7 ln(9x)}{7} - \frac{x^7}{49} +C \]
Determine whether f(x) is a probability density function on the interval \([1, e^6]\). If not, determine the value of the definite integral.
\[ f(x) = \frac{1}{6x} \]
We need to confirm the following:
\(f(x) = \frac{1}{6x}\) is clearly non-negative for \(x>1\) since both 6 & x are positive in the interval. We can confirm this satisfied.
Now, we will calculate the area under f(x) in the following interval: \([1,e^6]\).
\[ \int_{1}^{e^6} = \frac{1}{6} ln(x) |_1^{e^6}\\ = \frac{1}{6} ln(e^6) - \frac{1}{6} ln(1)\\ = 1- 0 = 1 \] The area under f(x) in the interval of \([1,e^6]\) is equal to 1. In conclusion, \(f(x) = \frac{1}{6x}\) is a probability density function on the interval \([1,e^6]\).