1.
Use integration by substitution to solv the integral below.
\[\int 4e^{-7x} dx\]
solution
\[u = -7x \\ \int 4e^u dx\\ \frac{du}{dx} = -7 \\ dx = \frac{du}{-7} \\ \int 4e^u \frac{du}{-7} \\ =\frac{4}{-7} \int e^u du \\ =\frac{4}{-7} e^u \\ = \frac{4}{-7}e^{-7x} + C\]
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 6530 bacteria per cubic centimeter.
solution
since \[\frac{dN}{dt} = -\frac{3150}{t^4} -220\]
then \[N(t) = \int -\frac{3150}{t^4} -220 dt \\ = -3150 \int \frac{1}{t^4} dt- \int 220 dt \\ = -3150 \int t^{-4} dt - \int 220 dt \\ = -3150 \frac{1}{-3t^3}-220t+C\]
after 1 day was 6530 bacteria per cubic centimeter, so \(t = 1\), \(N(1) = 6530\), plug these into \(N(t)\) function I got previously, I have
\[-3150 \frac{1}{-3*1^3}-220*1+C = 6530 \\ 1050-220+C = 6530\\ C = 5700\]
so, the complete function will be
\[N(t) = -3150 \frac{1}{-3*t^3}-220*t+5700\]
3.
Find the total area of the red rectangles in the figure below, where the equaltion of the line is \(f(x) = 2x-9\). (can’t copy the figure here)
observe the figure:
from the scale, each little grid is 1. the red regtangle starts from 4.5 to 8.5.
solution
\[\int_{4.5}^{8.5} 2x-9 dx \\ = \int 2x dx - \int 9 dx \\ = x^2 - 9x |_{4.5}^{8.5} \\ = (8.5^2 - 9*8.5) - (4.5^2 - 9 * 4.5)\\ = 16\]
so, the area = 16
4.
Find the area of the region bounded by the graphs of the given equations
\[y = x^2 -2x-2, y = x+2\]
solution
let \(f(x) = x^2-2x-2, g(x) = x+2\)
x = seq(-10, 10,by = 0.001)
cal_fx = function(x1){
return(round(x1^2-2*x1-2, 2))
}
cal_gx = function(x2){
return(round(x2+2,2))
}
plot_graph = function(x, gx, fx){
plot(x, gx, 'l', col = 'blue', ylab = 'y')
lines(x, fx, col = 'green')
abline(h = 0, col = 'lightgrey')
title('f(x) and g(x)')
legend(-10,10, c('g(x)', 'f(x)'), lwd = c(1,1), col = c('blue', 'green'))
}
fx = cal_fx(x)
gx = cal_gx(x)plot \(f(x)\) and \(g(x)\) to make sure that they have interception, otherwise the bounded area will be ZERO. and \(g(x)\) is above \(f(x)\)
plot_graph(x,gx,fx)
calculate the interception of two lines,
\[x^2-2x-2=x+2\\ x^2-3x-4=0 \\ (x-4)(x+1)=0\\ x_1 = 4, x_2 = -1\]
As a result,
\[area = \int_{-1}^{4} (x+2)dx - \int_{-1}^{4} (x^2-2x-2)dx \\ = (\frac{1}{2}x^2+2x)|_{-1}^{4} - (\frac{1}{3}x^3-x^2-2x)|_{-1}^{4} \\ = \frac{125}{6}\]
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 number of orders per year that will minimize inventory costs.
solution
from the text, the inventory cost = cost of lot + cost of orders
let y = inventory cost,
let n = number of orders,
let x = lot size,
where nx = 110,
therefore,
\[y = 8.25n + 3.75\frac{x}{2} \\ = 8.25n+3.75\frac{\frac{110}{n}}{2} \\ = 8.25n+\frac{206.25}{n}\]
take the derivative and find the derivative is equal to zero,
\[\frac{dy}{dn} = 8.25-\frac{206.25}{n^2} = 0\]
n = 5, so x = 22, y = 82.5
6.
Use integration by parts to solve the integral below.
\[\int ln(9x) \cdot x^6 dx\]
solution
\[u = ln(9x) \\ du = \frac{1}{x}dx \\ dv = x^6 \\ v = \frac{x^7}{7}\]
formula for integration by parts: \(u\cdot v-\int v\cdot du\)
\[\int ln(9x) \cdot x^6 dx \\ =ln(9x) \cdot \frac{x^7}{7} - \int \frac{x^7}{7} \cdot \frac{1}{x}dx \\ = \frac{ln(9x) \cdot x^7}{7} - \frac{x^7}{7} \cdot \frac{1}{7}+C \\ =\frac{x^7}{7}\begin{bmatrix}ln(9x)-\frac{1}{7}\end{bmatrix} + C\]
7.
Determine whether \(f(x)\) is a probability density function on the interval \([1, e^6]\). If not, determine the value of the integral.
\[f(x) = \frac{1}{6x}\]
solution
\(x \not= 0\) to make the function valid
\[\int_{1}^{e^6} \frac{1}{6x}dx \\ = \frac{ln(x)}{6}|_{1}^{e^6} \\ = \frac{ln(e^6)}{6} - \frac{ln(1)}{6} \\ = 1-0 \\ = 1\]
the function converges on given interval, therefore, \(f(x)\) is a probability density function on the interval