\(\int 4e^{-7x} \, dx\)
Using U substitution
\[ \int 4e^{u} \, dx \]
\[ u = -7x \]
\[ du = -7dx \]
\[ dx = -\frac{du}{7} \]
Substituting the variables \[ \int 4e^{u} \, dx \]
\[ \int 4e^{u} \, \left(-\frac{du}{7}\right) \]
\[ -\frac{4}{7} e^{-7} + C \]
To find the function \(N(t)\), we integrate the rate of change of contamination with respect to time: \[ \int \frac{dN}{dt} \, dt = \int (-3150t^4 - 220) \, dt \]
Integrating both sides: \[ N(t) = -\frac{3150}{5}t^5 - 220t + C \]
The level after 1 day is 6530 bacteria per cubic centimeter, we can use this information to find the value of the constant \(C\): \[ N(1) = -\frac{3150}{5}(1)^5 - 220(1) + C = 6530 \]
Solving for \(C\): \[ -315 - 220 + C = 6530 \\ C = 6530 + 315 + 220 \\ C = 7065 \]
Therefore, the function \(N(t)\) to estimate the level of contamination is: \[ N(t) = -\frac{3150}{5}t^5 - 220t + 7065 \]
Total Area = _{4.5}^{8.5} (2x - 9) , dx
To find the area, we will integrate
\[ Total Area = \int_{4.5}^{8.5} (2x - 9) \, dx \] \[ Total Area = [x^2 - 9x]_{4.5}^{8.5} \] \[ Total Area = [(8.5)^2 - 9(8.5)] - [(4.5)^2 - 9(4.5)] \] \[ Total Area = −4.25 + 20.25 \] \[ Total Area = 16 \]
\[ x^2 - 2x - 2 = x + 2x^2 - 3x - 4 = 0 \]
\[ (x - 4)(x + 1) = 0 \]
\[ x = 4 \quad \text{and} \quad x = -1 \]
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:
\[ A = \int_{a}^{b} [\text{Top} - \text{Bottom}] \, dx = \int_{4}^{-1} (x + 2 - (x^2 - 2x - 2)) \, dx = \int_{4}^{-1} (-x^2 + 3x + 4) \, dx \]
\[ = \left[ -\frac{x^3}{3} + \frac{3}{2}x^2 + 4x \right]_{4}^{-1} = \left[ -\frac{64}{3} + 3(4)^2 + 4(4) \right] - \left[ -\frac{(-1)^3}{3} + \frac{3}{2}(-1)^2 + 4(-1) \right] \]
\[ = \left[ -\frac{64}{3} + 24 + 16 \right] - \left[ -\frac{1}{3} + \frac{3}{2} - 4 \right] = 20.8333 \]
The number of orders (x) multiplied by the lot size (n) equals 110.
Let C represent costs.
\[ C = 8.25x + \frac{3.75 \times 110}{x^2} \] \[ C = 8.25x + \frac{3.75 \times 55}{x} \] \[ C = 8.25x + 206.25x \]
To minimize costs, we set the derivative equal to zero.
\[ \frac{dC}{dx} = 8.25 - \frac{206.25}{x^2} = 0 \] \[ 0 = 8.25 - \frac{206.25}{x^2} \]
Solving for x, we get:
\[ 206.25x^2 = 8.25 \] \[ \frac{206.25}{8.25} = x^2 \] \[ 25 = x^2 \]
\(\int ln( 9x ) · x^6dx\)
Using integration by parts \[ \int u \, dv = uv - \int v \, du \] \[ u = \ln(9x) \quad v = \frac{x^7}{7} \] \[ du = \frac{1}{x} \, dx \quad dv = x^6 \, dx \]
\[ \int \ln(9x) \cdot x^6 \, dx = \ln(9x) \cdot \frac{x^7}{7} - \int \left(\frac{x^7}{7}\right) \cdot \frac{1}{x} \, dx \] \[ = \ln(9x) \cdot \frac{x^7}{7} - \int \frac{x^6}{7} \, dx \] \[ = \ln(9x) \cdot \frac{x^7}{7} - \frac{x^7}{49} + C \]
\(f( x ) = 1/6x\)
\[ ( f(x) = 16x ) \]
The integral of a PDF must be equal to 1:
\[ \int_{-\infty}^{\infty} e^{61} f(x) \, dx = \int_{-\infty}^{\infty} e^{61} 16x \, dx \]
\[ \int_{-\infty}^{\infty} e^{61} f(x) \, dx = 16 \int_{-\infty}^{\infty} e^{61} x \, dx \]
\[ \int_{-\infty}^{\infty} e^{61} f(x) \, dx = 16 \int_{-\infty}^{\infty} e^{61} x \, dx \]
\[ \int_{-\infty}^{\infty} e^{61} f(x) \, dx = 16 \int_{-\infty}^{\infty} e^{61} x \, dx \]
\[ \int_{-\infty}^{\infty} e^{61} f(x) \, dx = 16 \ln(x) \bigg|_{e^{61}} \]
\[ \int_{-\infty}^{\infty} e^{61} f(x) \, dx = 16 (\ln(e^{6}) - \ln(1)) \]
\[ \int_{-\infty}^{\infty} e^{61} f(x) \, dx = 16 \]
When we compute the definite integral of the function \(f(x) = 16x\) over the interval [0, \(e^6\)], the result is 1. Consequently, we conclude that the function \(f(x)\) acts as a probability density function.
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