Using integration by substitution to solve the integral: \[ \int 4e^{-7x} \, dx \]
Let \(u = -7x\), therefore \(du = -7dx\) or \(dx = -\frac{1}{7}du\). Substituting into the integral, we get: \[ \int 4e^{u} \cdot \frac{du}{-7} = -\frac{4}{7} \int e^{u} \, du = -\frac{4}{7}e^{u} + C \] Now, since \(u = -7x\), we replace \(u\) back with \(-7x\): \[ -\frac{4}{7} e^{u} + C = -\frac{4}{7} e^{-7x} + C \]
Thus, the integral \(\int 4e^{-7x} \, dx\) evaluates to: \[ \boxed{-\frac{4}{7} e^{-7x} + C} \]
To solve for the function \(N(t)\) that estimates the level of contamination in a pond, we start by setting up the differential equation based on the given rate of change of contamination. The rate of change \(dN/dt\) is described as follows:
\[ \frac{dN}{dt} = -\frac{3150}{t^4} - 220 \]
where \(t\) is the number of days since the treatment began.
The function \(N(t)\) can be found by integrating the rate of change:
\[ N(t) = \int \left(-\frac{3150}{t^4} - 220\right) \, dt \]
This integral can be computed as follows:
\[ N(t) = \int -\frac{3150}{t^4} \, dt - \int 220 \, dt \]
Integrating term-by-term, we get:
\[ N(t) = 1050 \frac{1}{t^3} - 220t + C \]
where \(C\) is the integration constant.
It is given that after 1 day the level of contamination was 6530 bacteria per cubic centimeter. This initial condition helps us determine \(C\):
\[ 1050 \cdot \frac{1}{1^3} - 220 \cdot 1 + C = 6530 \]
\[ 1050 - 220 + C = 6530 \]
\[ C = 6530 - 1050 + 220 = 5700 \]
Thus, the function that models the level of contamination \(N(t)\) is:
\[ N(t) = 1050 \frac{1}{t^3} - 220t + 5700 \]
To find the total area of the red rectangles under the line described by the function \(f(x) = 2x - 9\) from \(x = 4.5\) to \(x = 8.5\), we calculate the definite integral of the function over this range. The integral will give us the area between the line and the x-axis.
Function Definition: The line equation is given by: \[ f(x) = 2x - 9 \] We need to integrate this function from \(x = 4.5\) to \(x = 8.5\).
Calculating the Integral: The integral of \(f(x)\) from \(x = 4.5\) to \(x = 8.5\) is: \[ \int_{4.5}^{8.5} (2x - 9) \, dx \] Solving this integral: \[ \int (2x - 9) \, dx = x^2 - 9x + C \] Applying the limits of integration: \[ \left[ x^2 - 9x \right]_{4.5}^{8.5} = (8.5^2 - 9 \times 8.5) - (4.5^2 - 9 \times 4.5) \] \[ = (72.25 - 76.5) - (20.25 - 40.5) \] \[ = -4.25 + 20.25 = 16 \]
To find the area of the region bounded by the curves given by the equations \(y = x^2 - 2x - 2\) and \(y = x + 2\), we will follow these steps:
Setting the Equations Equal to Find Intersection Points: \[ x^2 - 2x - 2 = x + 2 \] solving for \(x\): \[ x^2 - 3x - 4 = 0 \] Factorizing the quadratic equation: \[ (x - 4)(x + 1) = 0 \] This gives the solutions \(x = 4\) and \(x = -1\).
Determining which Function is Above the Other: Substituting values between \(x = -1\) and \(x = 4\) into both functions to determine which is above the other. We observed that \(y = x + 2\) lies above \(y = x^2 - 2x - 2\) in this interval.
Setting Up the Integral for the Area:
Here, \(f(x) = x + 2\) and \(g(x) = x^2 - 2x - 2\). \[ A = \int_{-1}^{4} [(x + 2) - (x^2 - 2x - 2)] \, dx = \int_{-1}^{4} [-x^2 + 3x + 4] \, dx \]
Computing the Integral: \[ A = \left[ -\frac{1}{3}x^3 + \frac{3}{2}x^2 + 4x \right]_{-1}^{4} \] \[ A = \left( -\frac{1}{3}(4)^3 + \frac{3}{2}(4)^2 + 4(4) \right) - \left( -\frac{1}{3}(-1)^3 + \frac{3}{2}(-1)^2 + 4(-1) \right) \] Simplifying to get: \[ A = \left( -\frac{64}{3} + 24 + 16 \right) - \left( \frac{1}{3} + \frac{3}{2} - 4 \right) \] \[ A = -\frac{64}{3} + 40 + \frac{1}{3} - \frac{3}{2} + 4 \] \[ A = -\frac{64}{3} + \frac{1}{3} - \frac{4.5}{3} = -\frac{64 + 4.5 - 1}{3} + 44= -\frac{67.5}{3} = 40 -22.5 \]
Thus,
\[
Area = 21.5
\]
To determine the optimal lot size and number of orders per year that minimize inventory costs for a beauty supply store, we use the Economic Order Quantity (EOQ) model.
The EOQ model is defined by the following parameters and formula:
Annual Demand (\(D\)): The total units demanded over the course of a year.
Holding Cost (\(H\)): The cost to hold one unit in inventory for a year.
Ordering Cost (\(S\)): The fixed cost associated with placing an order.
The formula for EOQ is given by: \[ EOQ = \sqrt{\frac{2DS}{H}} \] Where: - \(D\) is the annual demand, - \(S\) is the ordering cost per order, - \(H\) is the holding cost per unit per year.
Annual demand \(D = 110\) flat irons.
Ordering cost \(S = \$8.25\) per order.
Holding cost \(H = \$3.75\) per unit per year.
# Define the parameters
D <- 110 # Annual demand
S <- 8.25 # Ordering cost per order
H <- 3.75 # Holding cost per unit per year
# Calculate EOQ
EOQ <- sqrt((2 * D * S) / H)
EOQ## [1] 22number_of_orders <- D / EOQ
number_of_orders## [1] 5The total inventory cost (TC) when ordering at the EOQ can be calculated as:
\[ TC = \left(\frac{D}{Q} \times S\right) + \left(\frac{Q}{2} \times H\right) \]
where \(Q\) is the order quantity, which is the EOQ in this case.
total_cost <- (D / EOQ) * S + (EOQ / 2) * H
total_cost## [1] 82.5Use integration by parts to solve the integral below: \[ \int x \ln(9x) \, dx \]
integration by parts to solve:
\[ \int \ln(9x) \cdot x \, dx \]
We choose: \[ u = \ln(9x), \quad dv = x \, dx \]
We integrate to find \(du\) and \(v\): \[ du = \frac{1}{x} \, dx, \quad v = \frac{x^2}{2} \]
Using the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \]
We substitute \(u\), \(dv\), \(du\), and \(v\): \[ \int \ln(9x) \cdot x \, dx = \frac{x^2}{2} \ln(9x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \] \[ = \frac{x^2}{2} \ln(9x) - \int \frac{x}{2} \, dx \] \[ = \frac{x^2}{2} \ln(9x) - \frac{x^2}{4} + C \]
Where \(C\) is a constant.
Determine whether the function \(f(x)\) is a probability density function on the interval \([1, e^6]\). If not, determine the value of the definite integral. The function is given by \[ f(x) = \frac{1}{6x}. \]
To determine whether the function \(f(x) = \frac{1}{6}x\) is a probability density function (PDF) on the interval \([1, e^6]\), we need to verify two conditions:
For \(f(x) = \frac{1}{6}x\), since \(x \geq 1\) over the interval \([1, e^6]\), \(f(x)\) is non-negative.
We calculate the integral of \(f(x)\) from 1 to \(e^6\):
\[ \int_{1}^{e^6} \frac{1}{6}x \, dx = \frac{1}{6} \int_{1}^{e^6} x \, dx \]
This integrates to:
\[ = \frac{1}{6} \left[\frac{x^2}{2}\right]_{1}^{e^6} = \frac{1}{6} \left(\frac{(e^6)^2}{2} - \frac{1^2}{2}\right) = \frac{1}{6} \left(\frac{e^{12}}{2} - \frac{1}{2}\right) \]
\[ = \frac{e^{12} - 1}{12} \]
Since this value is not equal to 1, \(f(x)\) is not a PDF.
To correct \(f(x)\), we need to find a constant \(c\) such that:
\[ c \int_{1}^{e^6} x \, dx = 1 \]
Solving for \(c\):
\[ c \cdot \frac{e^{12} - 1}{2} = 1 \Rightarrow c = \frac{2}{e^{12} - 1} \]
Thus, the corrected function is:
\[ \boxed{f(x) = \frac{2}{e^{12} - 1}x} \]
This function \(f(x)\) now qualifies as a PDF over the interval \([1, e^6]\).