Use integration by substitution to solve the integral below.
\(\int4e^{-7x} dx\)
Let u = -7x \(\int4e^{u} dx\)
\(\frac{du}{dx} = -7\) \(dx = \frac{du}{7}\)
\(\int{\frac4{7}e^u}du\)
\(\frac4{7} \int{e^u} du\)
\(\frac4{7}e^u\)
\(\frac4{7}e^{-7x} + C\)
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.
\(\int -3150 t^{-4} -220\)
\(-3150\frac{t^{-3}}{-3} -220t + C\)
\(6530 = -3150\frac{1^{-3}}{-3} -220(1) + C\) \(C = 6530 + 1270 = 7800\)
\(N(t) = -1050t^{-3} -220t + 7800\)
Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x) = 2x - 9\).
\(\int_{4.5}^{8.5} 2x-9 dx\)
\(\int_{4.5}^{8.5} x^2 -9x\)
\(((8.5)^2 - 9(8.5)) - ((4.5)^2 - 9(4.5)) = 16\)
Find the area of the region bounded by the graphs of the given equations. \(y = x^2 - 2x - 2, y = x + 2\)
\(A = \int{x^2 - 2x - 2 dx} - \int{x+2}\)
\(A = \int{x^2 -3x -4} dx\)
\((x+1),(x-4)\)
\(\int_{-1}^{4} x^2 -3x -4 dx\)
\(\frac{x^3}{3} - \frac{3x^2}{2} - 4x\)
\((\frac{64}{3} - \frac{48}{2} - 16) -( \frac{-1}{3} - \frac{3}{2} + 4) = -18.66667 - 2.166667 = -20.83334\)
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.
\(Storage_Cost = \frac{3.75n}{2} + \frac{8.25 * 110}{n}\)
\(f(x) = \frac{3.75n}{2} + \frac{8.25 * 110}{n}\)
\(f(x) = 1.875n + \frac{907.5}{n}\)
\(f'(x) = 1.875 + \frac{907.5}{n^2}\)
\(0 = 1.875 + \frac{907.5}{n^2}\)
\(1.875(n^2) = 907.5\)
\(n^2 = 484\)
\(n = 22\)
Use integration by parts to solve the integral below. \(\int{ln(9x)*x^6} dx\)
\(\int{udv} = uv - \int{vdu}\)
let \(u = ln(9x)\) and \(v = \frac{1}{7}x^7\), \(\frac{du}{dx} = \frac1{x}\) and $ = x^6$
\(\int{ln(9x)*x^6dx} = ln(9x)*\frac{1}{7}x^7 - \int{\frac{1}{7}x^7*\frac1{x}dx}\)
\(= ln(9x)*\frac{1}{7}x^7 - \int{\frac1{7}x^6}\)
\(= ln(9x)*\frac{1}{7}x^7 - \int{\frac1{7}x^6}\)
\(= ln(9x)*\frac{1}{7}x^7 - \frac1{49}x^7\)
\(= \frac1{49}x^7(7ln(9x)* - 1)\)
Determine whether \(f(x)\) is a probability density function on the interval \([1, e^6]\). If not, determine the value of the definite integral.
\(f(x) = \frac1{6x}\)
\(\int_{1}^{e^6} \frac1{6x} dx\)
\(\frac1{6} ln(x)\)
\(\frac1{6} ln(e^6)\) - \(\frac1{6} ln(1)\) = \(1 - 0 = 1\)
The function is a probability density function on the interval \([1, e^6]\), because the area encapsulated by the interval is 1.