Use integration by substitution to solve the integral below:
\[ \int 4 e^{-7x} dx \].
Let \(u = -7x\), therefore \(du = -7dx\).
The integral simplifies to: \[-\frac{4}{7} \int e^u du = -\frac{4}{7} e^{-7x} + C\]
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.
Initial condition is \(N(1) = 6530\):
\[\frac{dN}{dt} = -\frac{3150}{t^4} -220 \] \[dN = (-\frac{3150}{t^4} -220) dt\] \[N(t) +C_1 = \int(-\frac{3150}{t^4} -220) dt\] \[N(t) +C_1 = \frac{1050}{t^3} -220t + C_2\]
The simplified expression is as follows: \[N(t) = \frac{1050}{t^3} -220t + C\]
Given that \(N(1) = 6530\) \[N(1) = 1050 - 220 + C = 6530\] \[ C = 5700\] \[ N(t) = \frac{1050}{t^3} -220t + 5700 \]
Find the total area of the red rectangles in the figure below, where the equation of the line is \(f ( x ) = 2x - 9\).
This is a Riemann sum:
\[ A = \sum_{x=5}^8 f(x) \Delta x \]
N.B. \(\Delta x = 1\)
\[ A = \sum_{x=5}^8 (2x-9) = (2 \cdot 5 -9 ) +(2 \cdot 6 -9 ) + (2 \cdot 7 -9 ) + (2 \cdot 8 -9 ) = 16\]
Find the area of the region bounded by the graphs of the given equations. \(y = x ^2 - 2x - 2\) , \(y = x + 2\).
First lets find the intersection points, i.e. where : \[ x ^2 - 2x - 2 = x + 2 \]. \[ x ^2 - 3x - 4 = 0 \]. \[ (x + 1 )(x -4) = 0 \]
The intersection points are \(x = -1\) and \(x = 4\).
In this region the line \(y = x + 2\) is greater than \(y = x ^2 - 2x - 2\).
Therefore the area bounded by the two graphs can be expressed as:
\[ \int_{-1}^{4} | (x+2) - (x^2-2x-2) | dx \] \[ = \int_{-1}^{4} |-x^2+3x+4 | dx \]
Note that the quadratic equation \(-x^2+3x+4 = (-x+4)(x+1) >= 0\) for \(x \in [-1,4]\).
We can therefore drop the absolute value and simplify to:
\[ = \int_{-1}^{4} (-x^2+3x+4 ) dx \] \[ = -\frac{x^3}{3} + \frac{3 x^2}{2} + 4x \big|_{-1}^{4} = \frac{125}{6}\]
f = function(x) { return( (-(x^3)/3) + (3/2)*x^2 + 4*x) }
f(4) - f(-1)## [1] 20.83333A 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.
Let the function \(T(x)\) denote the total annual inventory cost, \(x\) denote the order quantity.
\[T(x)= 8.25\cdot \frac{110}{x} + 3.75 \cdot \frac{x}{2} \]
\[T'(x)= -8.25\cdot \frac{110}{x^2} + \frac{3.75}{2} \]
We will need to find the values of \(x\) for which \(T'(x) = 0\).
\[T'(x)= -8.25\cdot \frac{110}{x^2} + \frac{3.75}{2} = 0\]
\[ x^2 = \frac{110 \cdot 8.25 \cdot 2 }{3.75} = 484\]
Therefore, \(x = 22\) and the lot size \(n = \frac{110}{22}=5\).
Use integration by parts to solve the integral below:
\[ \int ln(9x) \cdot x^6 dx \]
Let \(u = ln(9x)\) , \(du = \frac{dx}{x}\).
Let \(dv = x^6 dx\), \(v = \frac{x^7}{7}\).
Integration by parts tell us that: \[ \int ln(9x) \cdot x^6 dx = \int u dv = uv - \int v du \] \[ = \frac{ln(9x)x^7}{7} - \int \frac{x^7}{7}\frac{dx}{x} \]
Assuming \(x \ne 0\)
\[ = \frac{ln(9x)x^7}{7} - \int \frac{x^6}{7} dx \] \[ = \frac{ln(9x)x^7}{7} - \frac{x^7}{49}+C \]
Determine whether \(f ( x )\) is a probability density function on the interval \([1, e^6]\) . If not, determine the value of the definite integral.
\[ f(x) = \frac{1}{6x}\]
If \(f ( x )\) then \(\int_1^{e^6} f(x) dx = 1\).
Consider the following: \[ \int_1^{e^6} f(x)dx = \int_1^{e^6} \frac{dx}{6x} = \frac{ln(x)}{6} \big|_1^{e^6} = 1 - 0 = 1\]
Also note, that \(f(x) > 0\) for \(x \in [1, e^6]\). It it therefore a probability density function on the interval.