\[ \int 4e^{-7x} \,dx \]
Let \(u = -7x\). \(du = -7dx \longrightarrow dx = -\frac {du}7\)
\[ \int 4e^{u} \,dx \]
From the rules of integration with exponential functions,
\[ \int -\frac 47 e^{u} \,du = -\frac 47 e^u + c\]
Finally, we substitute u out of the equation.
\[ \int 4e^{-7x} \,dx = -\frac 47 e^{-7x} + c\]
Pond Treatment:
\[\frac{dN}{dt} = \frac{3150}{t^4} - 220\]
Here, we need to solve the equation for \(dN\) and then integrate both sides of the equation.
\[dN = \frac{3150}{t^4} - 220 \, dt\]
\[\int dN = \int \frac{3150}{t^4} - 220 \, dt\]
\[N(t) = -\frac 13 \times\frac{3150}{t^3} - 220t + c\]
Next, we need to find the constant.
\[N(1)= 6530 = -\frac {3150}{3} - 220 + c\]
\[N(1)= 6530 = -\frac {3150}{3} - 220 + c\]
\[6530 = -1270 + c\]
\[c = 7800\]
Answer
\[N(t) = -\frac 13 \times\frac{3150}{t^3} - 220t + 7800\]
The rectangles span from x = 4.5 to x = 8.5.
\[ \int_{4.5}^{8.5} 2x - 9 \,dx = (8.5)^2 - 9(8.5) - ((4.5)^2 - 9(4.5))\]
\[ \int_{4.5}^{8.5} 2x - 9 \,dx = 16 \]
Thus, the area of the rectangles is 16.
\(f(x) = x^2 - 2x - 2, g(x) = x + 2\)
These two graphs intersect at \((4,6)\) and \((-1,1)\) and \(g(x)\) is the upper equation.
\[\int g(x) - f(x) \,dx = \int_{-1}^{4} x + 2 - (x^2 - 2x - 2)\, dx\]
\[\int g(x) - f(x) \,dx = \int_{-1}^{4} -x^2 + 3x + 4 \,dx\]
\[\int g(x) - f(x) \,dx = -\frac{x^3}{3} + \frac {3x^2}{2} + 4x\]
\[\int g(x) - f(x) \,dx = -\frac{4^3}{3} + \frac {3*(4)^2}{2} + 4*4 - (-\frac{(-1)^3}{3} + \frac {3(-1)^2}{2} + 4*(-1))\]
\[\int g(x) - f(x) \,dx = -\frac {64}{3} + 24 + 16 - \frac 13 - \frac 32 + 4\]
\[\int g(x) - f(x) \,dx = 20 \frac 56\]
We need to optimize the inventory cost by finding the minimum value of the derivative. The true minimum inventory cost would be to sell all of the flat irons at the beginning of the year and then close up shop. We need to find a realistic optimum value.
Assuming that the quantity in each order is fixed, the average amount of irons stored would be simply \(\frac{110}2\). The quantity of irons would start at the maximum and it would decrease linearly to 0. The curve would form a triangle with the x and y axes, hence the factor of \(\frac 12\).
s = storage cost n = number sold o = number of orders f = fixed cost c = total cost
Note, the number of irons in each order is equivalent to the total irons divided by the order size.
\(c = (\frac n2*s) + f*o\)
\(c = (\frac {110}{2o}*s) + f*o\)
\(c = \frac {110*3.75}{2o} + 8.25*o\)
Differentiate c with respect to o.
\(c' = -\frac{206.25}{o^2} + 8.25\)
Let \(c' = 0\) and solve for o, the number of orders.
\(0 = -\frac{206.25}{o^2} + 8.25\)
\(8.25 = \frac{206.25}{o^2}\)
\(o = \sqrt{\frac{206.25}{8.25}}\)
\(o \approx 5\)
The total number of orders for an optimal inventory cost is 5
\[\int ln({9x}) * x^6 \, dx\]
\(u = ln(9x)\), \(du = \frac 1x\, dx\) \(dv = x^6\, dx\), \(v = \frac {x^7}{7}\)
\(\int udv = uv - \int vdu\)
\[\int ln({9x}) * x^6 \, dx = ln(9x) * \frac {x^7}{7} - \int \frac {x^7}{7} \frac 1x\, dx \]
\[\int ln({9x}) * x^6 \, dx = ln(9x) * \frac {x^7}{7} - \int \frac {x^6}{7}\, dx \]
\[\int ln({9x}) * x^6 \, dx = ln(9x) * \frac {x^7}{7} - \frac {x^7}{49} \]
\[\int ln({9x}) * x^6 \, dx = (\frac {x^7}{7})*(ln(9x) - \frac {1}{7} \]
Probability Density Functions must cover an area of 1, no more, and no less.
\[\int_{1}^{e^6} \frac 1 {6x} \, dx = \left. \frac {ln(x)}{6} \right \rvert_{1}^{e^6} \]
\[\int_{1}^{e^6} \frac 1 {6x} \, dx = \frac {ln(e^6)}{6} - \frac{ln(1)}{6}\]
\[\int_{1}^{e^6} \frac 1 {6x} \, dx = \frac {6ln(e)}{6} - 0\]
Because \(ln(e) = 1\),
\[\int_{1}^{e^6} \frac 1 {6x} \, dx = 1\]
Thus, \(f(x) = \frac 1 {6x}\) is a probability density function on the interval \([1,e^6]\).