Integrate by substitution: \(\int{4e^{-7x} \ dx}\)
Let \(u\) = \(e^{-7x}\)
\(\frac{du}{dx} = -7e^{-7x}\)
\(du = -7e^{-7x} \ dx\)
\(\frac{-du}{7} = e^{-7x} \ dx\)
Substitute:
\(\int{4 \cdot \frac{-du}{7}}\)
\(\frac{-4}{7} \int{du}\)
\(\frac{-4}{7} \cdot u\)
\(\frac{-4}{7} \cdot e^{-7x}\)
\(\frac{dN}{dt} = \frac{-3150}{t^{4}} - 220\)
Integrate:
\(-3150 \cdot \frac{-1}{3} \cdot \frac{1}{t^{3}} - 220t + C\)
\(N(t) = \frac{1050}{t^{3}} -220t + C\)
After 1 day,
\(N(1) = \frac{1050}{(1)^{3}} - 220(1) + C = 6530\)
\(N(1) = 830 + C = 6530\)
\(C=5700\)
Therefore,
\(N(t) = 1050 \cdot \frac{1}{t^{3}} - 220t + 5700\)
\(A = 1 \cdot f(5) + 1 \cdot f(6) + 1 \cdot f(7) + 1 \cdot f(8)\)
\(A = 1 \cdot (2\cdot5-9) + 1 \cdot (2\cdot6-9) + 1 \cdot (2\cdot7-9) + 1 \cdot (2\cdot8-9)\)
\(A = 1+3+5+7\)
\(A = 16\)
\(\int_{-1}^{4} x+2-(x^{2}-2x-2) \ dx\)
= \(\int_{-1}^{4} (-x^{2}+3x+4) \ dx\)
= \(\int_{-1}^{4} (-x^{2}+3x+4) \ dx\)
= \(\frac{-x^{3}}{3} + \frac{3x^{2}}{2} + 4x |_{-1}^{4}\)
= \((\frac{-4^3}{3} + \frac{3(4^{2})}{2} + 4(4) ) - (\frac{-(-1^{3})}{3} + \frac{3(-1)^{2}}{2} + 4(-1))\)
= \((\frac{-64}{3} + \frac{48}{2} + 16) - (\frac{1}{3} + \frac{3}{2} -4)\)
= \(\frac{56}{3} - (\frac{-13}{6})\)
= \(\frac{125}{6}\)
Let:
The total inventory cost is:
\[ TC = OC + HC \]
The cost of placing orders is:
\[ OC = \frac{D}{Q} \times H \]
The cost of holding the inventory is:
\[ HC = \frac{Q}{2} \times S \]
We want to minimize the total inventory cost, so we need to minimize the sum of the ordering cost and the holding cost with respect to \(Q\).
\[ TC = \frac{D}{Q} \times H + \frac{Q}{2} \times S \]
Differentiate with respect to \(Q\) and set to zero.
\[ \frac{d(TC)}{dQ} = \frac{DH}{Q^2} + \frac{S}{2} = 0 \]
\[ \frac{DH}{Q^2} = - \frac{S}{2} \]
\[ Q^2 = \frac{2DH}{S} \]
\[ Q = \sqrt{\frac{2DH}{S}} \]
Plug in the given values:
\[ Q = \sqrt{\frac{2 \times 110 \times 8.25}{3.75}} \]
\[ Q \approx \sqrt{\frac{1815}{3.75}} \]
\[ Q \approx \sqrt{484} \]
\[ Q \approx 22 \]
So, the optimal lot size is approximately 22 flat irons per order.
The number of orders per year is:
\[ \frac{D}{Q} = \frac{110}{22} = 5 \]
The optimal number of orders per year is 5.
The beauty supply store should order approximately 22 flat irons per order and they should place 5 orders per year.
Use integration by parts to solve : \(\int ln(9x) \cdot x^{6} \ dx\)
Let \(u=ln(9x)\)
\(\frac{du}{dx} = \frac{1}{x}\)
\(du = \frac{dx}{x}\)
Let \(v = \int{x^{6} \ dx} = \frac{1}{7}x^{7}\)
\(\frac{dv}{dx}=\frac{6}{7}x^{6}\)
\(dv = \frac{6}{7}x^{6} \ dx\)
Substituting into the integration by parts formula,
\(\int {ln(9x) \frac{6}{7}x^{6} \ dx} = ln(9x) \cdot \frac{1}{7}x^{7} - \int {\frac{1}{7}x^{7} \cdot \frac{1}{x} \ dx}\)
\(\frac{x^{7}ln(9x)}{7} - \frac{1}{7} \int{x^{6} \ dx}\)
\(\frac{x^{7}ln(9x)}{7} - \frac{1}{7} \cdot \frac{x^{7}}{7}\)
\(\frac{x^{7}ln(9x)}{7} - \frac{x^{7}}{49}\)
\(f(x) = \frac{1}{6x}\)
For \(f(x)\) to be a probability density function within the interval \([1, e^{6}]\), it must integrate to 1 along this interval. It also must be non-negative along the interval.
Integrating:
\(\int_{1}^{e^{6}} \frac{1}{6x} \ dx\)
= \(\frac{1}{6} ln(e^{6}) - \frac{1}{6}ln(1)\)
= \(\frac{1}{6}ln(e^{6}) - 0\)
= \(\frac{1}{6} \cdot 6\)
= 1
Since \(f(x)\) is non-negative along the interval, and it integrates to 1, it is a probability density function along the interval.