Use integration by substitution to solve the integral below. \[\int 4e^{-7x}dx\]
Let \(u = -7x\), \(du = -7dx\), Then, \(dx=\frac{du}{-7}\) \[\int 4e^u\frac{du}{-7}\] \[\frac{4}{-7}\int e^udu\] \[\frac{4}{-7}e^u+c\] \[\therefore \frac{4}{-7}e^{-7x}+c\]
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 6530 bacteria per cubic centimeter.
\[N'(t)=\frac{dn}{dt}=\int(-\frac{3150}{t^4}-220)dt\] \[N(t)=\frac{1050}{t^3}-220t+C\] \[N(1)=1050-220+C=6530\] \[C=5700\] \[\therefore N(t)=\frac{1050}{t^3}-220t+5700\]
Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x)=2x-9\).
\[\int^{8.5}_{4.4}(2x-9)dx\] \[x^2-9xdx|^{8.5}_{4.5}\] \[\therefore ((8.5)^2-9*(8.5) - (4.5)^2-9*(4.5) = 16\]
Find the area of the region bounded by the graphs of the given equations. \[y=x^2-2x-2, y=x+2\]
\[x^2-2x-2=x+2\] \[x^2-3x-4=0\] \[x=1, x=-4\] \[\int_{-1}^{4}(x+2)-(x^2-2x-2)dx\] \[ \int_{-1}^{4}(-x^2+3x+4)dx\] \[-\frac{x^3}{3}+\frac{3x^2}{2}+4x|^{4}_{-1}\] \[\therefore (-\frac{4^3}{3}+\frac{3(4)^2}{2}+4(4))-(-\frac{-1^3}{3}+\frac{3(-1)^2}{2}+4(-1))=20.83333\]
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.
Let \(x\) be a number of flat irons,
Yearly storage cost = Storage cost per iron × Average number of irons = \(3.75.\frac{x}{2}=1.875x\)
Yearly ordering cost = Cost of each order × Number of orders = \(8.25.\frac{110}{x}=\frac{907.5}{x}\)
Inventory cost = Yearly storage cost + Yearly ordering cost = \(1.875x+\frac{907.5}{x}=f(x)\) \[f'(x)=1.875-\frac{907.5}{x^2}=0\] \[lot.size=x=22, order=\frac{110}{22}=5\]
The lot size is 22 and the number of orders per year of 5 will minimize inventory costs.
Use integration by parts to solve the integral below. \[\int ln(9x).x^6 dx\] Let \(u=ln(9x)\), then \(\frac{du}{dx}=\frac{1}{x}\)
Let \(\frac{dv}{dx}=x^6\), then \(v=\int x^6dx=\frac{1}{7}x^7\)
Using the formula for integration by parts: \(\int u\frac{dv}{dx}dx=uv-\int v\frac{du}{dv}dx\) \[\int ln(9x).x^6 dx=\frac{1}{7}x^7.ln(9x)-\int \frac{1}{7}x^7.\frac{1}{x}dx \] \[\frac{1}{7}x^7.ln(9x)-\int \frac{1}{7}x^6dx \] \[\frac{7}{49}x^7.ln(9x)-\frac{1}{49}x^7+C \] \[\frac{1}{49}(7ln(9x)-1)+C \]
Determine whether \(f(x)\) is a probability density function on the interval \([1, e^6]\). If not, determine the value of the definite integral. \[f(x)=\frac{1}{6x}\] \[\int^{e^6}_{1}\frac{1}{6x}dx=\frac{1}{6}ln(x)|^{e^6}_{1}\] \[\frac{1}{6}ln(e^6)-\frac{1}{6}ln(1)=1\] Since the definite integral is 1, f(x) is a probability density function on the interval \([1, e^6]\)