1.Use integration by substitution to solve the integral below.
\[\int 4e^{-7x}dx\]
\(u = -7x\)
\(du/dx = -7\)
\(du = -7dx\)
\[-4*\frac{1}{7}\int-7e^{-7x}dx=-\frac{4}{7}\int e^{u}du=-\frac{4}{7}e^{-7x} + c \]
2.Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \(\frac{dN}{dt}=−\frac{3150}{t^4}−220\) bacteria per cubic centimeter per day, where \(t\) is the number of days since treatment began. Find a function \(N(t)\) to estimate the level of contamination if the level after 1 day was 6350 bacteria per cubic centimer.
\[\int \frac{dN}{dt} = \int -3150t^{-4}-220dt\]
\[N(t) = \frac{3150}{3}t^{-3} - 220t + c = \frac{1050}{t^3} - 220t + c\]
\[N(1) = \frac{1050}{(1)^3} - 220(1) + c = 1050 - 220 + c = 830 + c=6350\]
\[c = 6350 - 830 = 5520\]
\[N(t) = \frac{1050}{t^3} - 220t + 5520\]
3.Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x) = 2x- 9.\)
\[F(x) = \int_{4.5}^{8.5} f(x) dx = \int_{4.5}^{8.5} 2x-9 dx = x^2-9x + c\bigg|_{4.5}^{8.5}\] \[[8.5^2-9(8.5)]-[4.5^2-9(4.5)]=-4.25+20.25=16\] 4.Find the area of the region bounded by the graphs of the given equations.
\[y = x^2 - 2x -2, y = x +2\] Solving the sytem of equeations gives us: \[x^2-2x-2=x+2\] \[x^2-3x-4=0\] \[(x-4)(x+1)=0; x=4, x=-1\]
\[\int_{-1}^{4}x^2-3x-4=\frac{x^3}{3}-\frac{3x^2}{2}-4x\bigg|_{1}^{4}=\bigg[\frac{4^3}{3}-\frac{3(4)^2}{2}-4(4)\bigg]-\bigg[\frac{-1^3}{3}-\frac{3(-1)^2}{2}-4(-1)\bigg]\]
\[=\bigg[\frac{64}{3}-\frac{48}{2}-16\bigg]-\bigg[\frac{-1}{3}-\frac{3}{2}+4\bigg]= 20.83\]
5.A beauty supply store expects to sell 110 flat irons during the next year. It costs $3.75 to store one flat iron for one year. There is a fixed cost of $8.25 for each order. Find the lot size and the number of orders per year that will minimize inventory costs.
c = cost
r = orders per year
x = irons
rx = 110
r = 110/x
\[c = 8.25r + \frac{375}{2r} \]
\[c^\prime=8.25-\frac{206.25}{r^2}=0\]
\[r=\sqrt{\frac{206.25}{8.25}}=5\] \[x = \frac{110}{r} = \frac{110}{5} = 22\]
\[c = 8.25(5)+\frac{375}{2(5)}=78.75\] The store should place orders in lots of 22 five times per year, keeping inventory costs at $78.75.
6.Use integration by parts to solve the integral below.
\[\int{ln(9x)*x^6dx}\]
\(u=ln(9x), \frac{dv}{dx}=x^6\)
\(du=\frac{1}{x}dx\)
\(dv=x^6dx\)
\(v=\frac{1}{7}x^7\)
\[\int{udv}=uv-\int{vdu}\] \[ln(9x)*\frac{1}{7}x^7-\int{\frac{1}{7}x^7*\frac{1}{x}dx}=ln(9x)\frac{x^7}{7}-\frac{x^7}{49}-c\]
7.Determine whether \(f(x)\) is a probability density function on the interval \(1, e^6\) . If not, determine the value of the definite integral.
\[f(x)=\frac{1}{6x}\]
\[g(x)=\int_{1}^{e^6}\frac{1}{6x}dx\]
\[\frac{1}{6}\int_{1}^{e^6}\frac{1}{x}dx=\frac{1}{6}ln(x)\bigg|_{1}^{e^6}\]
\[\bigg[\frac{1}{6}ln(e^6)\bigg]-\bigg[\frac{1}{6}ln(1)\bigg]=\frac{1}{6}*6-0=1\]
The function is a probabilitiy density function.