First move the constant out front, \[ = 4 * \int{e^{-7x}dx} \] Substitute u, \[ u = -7x \rightarrow \frac{du}{dx}= -7 \rightarrow du = -7dx \rightarrow dx = -\frac{1}{7}du \] \[ = 4 * \int{e^udx} \rightarrow e^u-\frac{1}{7}du \rightarrow -\frac{1}{7}e^udu \] \[ = 4 * \int{-\frac{1}{7}e^udu} \]

Since \[\int{e^udu} = e^u\]

\[= 4(-\frac{1}{7}e^u)\] and \[e^u = -7x\] so

\[= 4(-\frac{1}{7}e^{-7x}) \rightarrow -\frac{4}{7}e^{-7x}\] Add a constant \(C\) to the solution,

\[= -\frac{4}{7}e^{-7x} + C\]

Given \(t = 1\) and \(N(1) = 6530\),

\(\frac{dN}{dt} = -\frac{3150}{t^4}-220 = -3150t^{-4}-220\)

\(N( t ) = \int{-3150t^{-4}-220 dt}\)

\(= \frac{-3150t^{-3}}{-3} - 220t + C\)

\(N(t) = \frac {1050}{t^3} - 220t + C\)

Since \(N(1) = 6530\),

\(6530 = \frac {1050}{1^3} - 220(1) + C\)

\(\Rightarrow 6530 = 1050 - 220 + C\) \(\Rightarrow 5700 = C\)

So,

\(N(t) = \frac {1050}{t^3} - 220t + 5700\)

We’re looking for the rectangle that starts at x = 4.5 and ends at x = 8.5

\(\int_{4.5}^{8.5} 2x - 9dx\)

\(=\left. \ {x^2} -9x \right|_{4.5}^{8.5}\)

Area = \((8.5^2 - 9(8.5)) - (4.5^2 - 9(4.5)) \Rightarrow 16\)

\(x^2 - 2x -2 = x + 2 \Rightarrow\)
\(x^2 - 3x - 4 = 0 \Rightarrow\) \((x - 4)(x + 1)\)

\(x = 4, x = -1\)

\(\int_{-1}^{4} (x+2)-(x^2-2x-2)dx \Rightarrow\)
\(\int_{-1}^{4} -x^2+3x+4dx\)

\(=\left. \ \frac{-x^3}{3}+ \frac{3x^2}{2} +4x \right|_{-1}^{4}\)

Area = \((\frac{-4^3}{3}+ \frac{3*(4)^2}{2} +4(4)) - (\frac{-(-1)^3}{3}+ \frac{3*(-1)^2}{2} +4(-1)) \Rightarrow\)

\(\frac{125}{6} = 20.83\)

y1 = function(x){x^2 - (2*x) - 2}
y2 = function(x){x + 2}

int_1 = integrate(y1, lower = -1, upper = 4)
int_2 = integrate(y2, lower = -1, upper = 4)

area = int_2$value - int_1$value
area
## [1] 20.83333

Let \(x\) = number of irons per order.

Then \(n\) = number of sales per year (110)

Then \(\frac{n}{x}\) is the number of orders per year (\(\frac{110}{x}\))

Annual storage cost is \(3.75 * \frac{x}{2} = 1.875x\)
Annual ordering cost is \(8.25 * \frac{110}{x} = \frac{907.5}{x}\)
Annual inventory cost \(f(x)\) is annual storage cost + annual ordering cost \(\Rightarrow 1.875x + \frac{907.5}{x}\)

To find the minimum value, find \(f'(x) = 0\)

\(f(x) = 1.875x + \frac{907.5}{x} \Rightarrow\)

\(f'(x) = 1.875 -\frac{907.50}{x^2}\)
\(0 = 1.875 -\frac{907.50}{x^2}\)
\(\frac{907.50}{x^2} = 1.875\)
\(x = 22\)

Since the ideal lot size is \(\frac{n}{x}\), the ideal size is \(\frac{110}{22} \Rightarrow 5\), and the ideal number of orders is 22.

Let \(u = ln(9x)\), then \(\frac{du}{dx} = \frac{1}{x}\) and \(du = \frac{1}{x}dx\),
Let \(dv = x^6\), then \(v = \frac{1}{7}x^7\)

Integrating,

\(uv-\int vdu \Rightarrow\)

\(\frac{x^{7}ln(9x)}{7}-\int \frac{1}{7}x^{7}\frac{1}{x}dx\)
\(\frac{x^{7}ln(9x)}{7}-\frac{1}{7} \int \frac{x^{7}}{x}dx\)
\(\frac{x^{7}ln(9x)}{7}-\frac{1}{7}\int x^{6}dx\)
\(\frac{x^{7}ln(9x)}{7}-\frac{1}{7}(\frac{x^{7}}{7})+C\)
\(\frac{x^{7}ln(9x)}{7}-\frac{x^{7}}{49}+C\)

\(\int_1^{e^6}\frac{1}{6x} dx\)
\(= \frac{1}{6} ln(x)|_1^{e^6}\)
\(= \frac{1}{6} ln(e^6) - \frac{1}{6} ln(1)\)
\(= \frac{1}{6} * 6 - \frac{1}{6} * 0\)
\(= 1\)

The definite integral is 1, so it is on the interval \([1, e^6]\)