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\[ 1.\ Integrate\ \int 4e^{-7x} dx\\ \\ Let\ u\ =\ -7x\ \Longrightarrow \ dx\ =\ -\frac{du}{7}\\ \\ We\ have:\int 4e^{-7x} dx\ =\ 4\int \frac{e^{u}}{-7} du\ =\ -\frac{4}{7} e^{u} \ +\ c\ =\ -\frac{4}{7} e^{-7x} +\ C\\ \\ \\ \] \[ 2.\ Rate\ of\ change\ and\ estimation\ function.\\ \\ \frac{dN}{dt} =-\frac{3150}{t^{4}} -220\\ \\ dN\ =\left( \ -\frac{3150}{t^{4}} -220\right) dt\\ \\ N\ =\ \int \left( \ -\frac{3150}{t^{4}} -220\right) dt\\ \\ N\ =\ \frac{1050}{t^{3}} -220t\ +\ C\\ N( 1) \ =\ 6530\ \Longrightarrow \ 6530\ =\ 1050-220+C\\ \Longrightarrow \ C\ =\ 5700\\ \\ Therefore,\ N( t) \ =\ \ \frac{1050}{t^{3}} -220t\ +\ 5700\\ \\ \\ \\ \] \[ 3.\ Area\ of\ red\ recatngle:\\ \\ f( x) \ =\ 2x-9\\ \\ The\ line\ is\ bounded\ at\ x\ =\ 4.5\ and\ x\ =\ 8.5\\ \\ Let\ find\ the\ area:\\ \\ Area\ =\ \int _{4.5}^{8.5}( 2x-9) dx\ =\ x^{2} \ -\ 9x\ |\ x\ =\ 4.5,\ 8.5\\ \\ =\ \left( 8.5^{2} -9( 8.5)\right) -\left( 4.5^{2} -9( 4.5)\right)\\ \\ =\ -4.25-( -20.25) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \\ =\ -4.25+20.25\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \\ =\ 16\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \\ \\ \] 4. Area under the curve Plotting the two functions…

f1 <- function(x) x^2 - 2*x - 2
f2 <- function(x) x + 2
x1 <- seq(-1.5, 6, 1)
plot(x1, f1(x1), type ='l', col='blue', xlab = "", ylab = "")
lines(x1, f2(x1), col='red')
abline(h=0)

\[ y\ =\ x^{2} -2x-2,\ y\ =\ x+2\\ \\ Findind\ the\ area\ boundaries:\ x^{2} -2x-2\ =\ x+2\\ \Longrightarrow \ x^{2} -3x-4\ =0\ \Longrightarrow \ x\ =\ -1,\ x=4\\ \\ Area\ =\ \int _{-1}^{4}\left(( x+2) -\left( x^{2} -2x-2\right)\right) dx\\ \\ =\ \int _{-1}^{4}\left( -x^{2} +3x+4\right) dx\ =\ -\frac{x^{3}}{3} +\frac{3}{2} x^{2} +4x\ |x=-1,\ 4\\ =\ \left( -\frac{64}{3} +24+16\right) -\left(\frac{1}{3} +\frac{3}{2} -4\right)\\ =\ \frac{56}{3} +\frac{13}{6}\\ \\ =\ \frac{112+13}{6}\\ \\ =\frac{125}{6}\\ \\ \\ \] \[ 5.\ \ Minimize\ the\ cost\\ \\ We\ know\ that:\ Total\ cost\ =\ fixed\ cost\ +\ variable\ cost\\ \\ Let\ x\ be\ number\ of\ order\\ \\ fixed\ cost\ =8.25x\\ \\ Variable\ cost\ is\ function\ of\ lot\ size\ as\ well.\\ \\ lot\ size\ =s\ =\ \frac{total\ of\ irons\ sold}{number\ of\ order}\\ \\ Thus,\ s\ =\ \frac{110}{x}\\ \\ Cost\ over\ the\ year\ \Longrightarrow \ uniform\ distribution\ of\ lot\ size,\ s\ =\ s/2:\\ \\ We\ can\ define\ the\ total\ cost\ function:\ C( x) \ =\ 8.25x\ +\ 3.75\left(\frac{\frac{110}{x}}{2}\right)\\ \\ \Longrightarrow \ C( x) \ =\ 8.25x\ +\ \frac{206.25}{x}\\ \\ First\ derivative\ to\ minimize\ the\ cost:\ C'( x) \ =\ 8.25-\frac{206.25}{x^{2}}\\ C'( x) \ =\ 0\ \Longrightarrow \ x^{2} \ =\ \frac{206.25}{8.25} \ \Longrightarrow \ x\ =\ 5\\ \\ x\ =\ 5\ \Longrightarrow \ s\ =\ \frac{110}{5} =22\\ \\ \\ The\ lot\ size\ and\ the\ number\ of\ orders\ per\ year\ that\ will\ minimize\ inventory\ costs\\ \\ are\ respectively\ 22\ and\ 5.\ \\ \\ \] \[ 6.\ Integrate\ \ by\ parts:\ \int ln( 9x) x^{6} dx\\ \\ We\ know\ that:\ \int udv\ =\ uv-\int vdu\\ \\ du\ =\ x^{6} dx\ \Longrightarrow \ u\ =\int x^{6} dx\ \Longrightarrow \ u\ =\ \frac{x^{7}}{7}\\ v\ =\ ln( 9x) \ \Longrightarrow \ dv\ =\frac{1}{x} dx\\ \\ So:\ \int ln( 9x) x^{6} dx\ =\int vdu\ \ =uv-\int udv=\left(\frac{x^{7}}{7}\right)( ln( 9x) -\int \frac{x^{7}}{7}\left(\frac{1}{x}\right) dx\\ \\ =\ \frac{1}{7} x^{7} ln( 9x) -\frac{1}{7}\int x^{6} dx\\ \\ =\ \ \frac{1}{7} x^{7} ln( 9x) -\frac{1}{7}\left(\frac{x^{7}}{7}\right) \ +\ C\\ \\ =\ \ \ \frac{1}{7} x^{7} ln( 9x) -\frac{1}{49} x^{7} +C\\ \\ \\ \\ \] \[ 7.\ Probability\ function\\ \\ We\ need\ to\ verify\ if\ \int _{1}^{e^{6}} f( x) dx\ =\ 1.\\ \\ \\ We\ have:\ \int _{1}^{e^{6}} f( x) dx\ \ =\ \int _{1}^{e^{6}}\frac{1}{6x} dx\ =\ \frac{1}{6} ln( x) \ |\ x\ =1,\ e^{6}\\ \\ =\ \frac{1}{6} ln\left( e^{6}\right) -\frac{1}{6} ln( 1)\\ \\ =1-0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \\ =\ 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \\ \\ \\ Since\ \ \int _{1}^{e^{6}} f( x) dx\ =\ 1,\ we\ can\ \ conclude\ that\ \ f( \ x\ ) \ is\ a\ probability\ density\ function\ \\ on\ the\ interval\ \left[ 1,\ e^{6} \ \right]\\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \\ \]