Use integration by substitution to solve the integral below.
\(\int { 4{ e }^{ -7x }dx }\)
Solution:
Let u = -7x, then du = -7dx. In terms of dx, dx = \(-\frac { 1 }{ 7 } du\)
Rewrite integral in terms of u.
\(\int { 4{ e }^{ u }(-\frac { 1 }{ 7 }) du }\) Take out constants.
\(-\frac { 4 }{ 7 }\int { { e }^{ u } du }\) Integrate.
\(-\frac { 4 }{ 7 }({ e }^{ u } + C)\) Substitute back in terms of x.
\(-\frac { 4 }{ 7 }({ e }^{ -7x } + C)\) For some constant 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.
Solution:
\(\int { -\frac { 3150 }{ { t }^{ 4 } } -220 }dt\) Split integral.
\(\int { -\frac { 3150 }{ { t }^{ 4 } }dt -\int { 220 }dt }\) Take out constants for both integrals
\(-3150\int { \frac { 1 }{ { t }^{ 4 }dt } -\int { 220 }dt }\) Integrate.
\(-3150(-\frac { 1 }{ 3{ t }^{ 3 } } )-220t\) Simplify.
\(\frac { 1050 }{ { t }^{ 3 } } -220t\) Write as function N(t).
\(N(t)=\frac { 1050 }{ { t }^{ 3 } } -220t+C\) For some constant C.
Solve for N(t) = 6530 where t = 1.
\(\frac { 1050 }{ { t }^{ 3 } } -220t+C=6530\)
\(\frac { 1050 }{ { 1 }^{ 3 } } -220(1)+C=6530\)
\(1050-220+C=6530\)
\(830+C=6530\)
\(C=5700\)
Function 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.
Solution:
From looking at the graph, it looks like there are 4 red rectangles that start at 4.5 and end at 8.5.
function_3 <- function(x)
{
2 * x - 9
}
integrate(function_3, lower = 4.5, upper = 8.5)## 16 with absolute error < 1.8e-13The total area of the red rectangles is 16.
Find the area of the region bounded by the graphs of the given equations.
\(y={ x }^{ 2 }-2x-2,y=x+2\)
Enter your answer below.
Solution:
Find the intersection of the given equations
\({ x }^{ 2 }-2x-2=x+2\) Set equal to 0.
\({ x }^{ 2 }-3x-4=0\) Factor.
\((x-4)(x+1)=0\) Solve for x.
\(x=4,x=-1\)
Subtract the two equations.
\(x+2-({ x }^{ 2 }-2x-2)\) Simplify.
\(x+2-{ x }^{ 2 }+2x+2\) Combine like terms.
\(-{ x }^{ 2 }+3x+4\)
Integrate this equation using intersection points as interval.
function_4 <- function(x)
{
-x ^ 2 + 3 * x + 4
}
integrate(function_4, lower = -1, upper = 4)## 20.83333 with absolute error < 2.3e-13The area of the region bounded by the graphs of the given equations is 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.
Solution:
The lot size will be 110 / the number of orders per year. Let us have n = number of orders per year and lot size = l. Then l = 110 / n. Our cost function will look like \(8.25n\quad +\quad 3.75(\frac { l }{ 2 } )\) This function simplified is
\(8.25n + 3.75(\frac { l }{ 2 } )\) Plug in l.
\(8.25n + 3.75(\frac { \frac { 110 }{ n } }{ 2 } )\) Simplify fraction.
\(8.25n+(\frac { 110*3.75 }{ 2n } )\) Cost function is equal to.
\(8.25n-(\frac { 206.25 }{ n } )\)
Take the derivative of the cost function.
\(8.25n-(\frac { 206.25 }{ n } )\) derivative =
\(8.25-(\frac { 206.25 }{ { n }^{ 2 } } )\)
Solve for n by setting derivative of cost function equal to 0.
\(8.25-(\frac { 206.25 }{ { n }^{ 2 } } )=0\) Move 8.25 to other side.
\(-(\frac { 206.25 }{ { n }^{ 2 } } )=-8.25\) Cross multiply.
\(-8.25{ n }^{ 2 }=-206.25\) Divide -8.25.
\({ n }^{ 2 }=\frac { -206.25 }{ -8.25 }\) Simplify.
\({ n }^{ 2 }=25\) Solve for n.
\(n=5\)
Now that we know n we can solve for l.
\(l=\frac { 110 }{ n }\) Plug in 5 for n.
\(l=\frac { 110 }{ 5 }\) Simplify.
\(l=22\)
The lot size and the number of orders per year that will minimize inventory costs are a lot size of 22 and the number of orders per year equal to 5.
Use integration by parts to solve the integral below.
\(\int { ln(9x)·{ x }^{ 6 }dx }\)
Solution:
Let u = ln(9x), then du = \(\frac { 1 }{ x } dx\). Let v = \(\frac { 1 }{ 7 } { x }^{ 7 }\), then dv = \({ x }^{ 6 }\).
Using integration by parts, we get u * v - the integral of v * du.
\(\ln { (9x)(\frac { 1 }{ 7 } { x }^{ 7 })-\int { \frac { 1 }{ 7 } { x }^{ 7 }(\frac { 1 }{ x } dx) } }\) Simplify.
\(\frac { { x }^{ 7 }ln(9x) }{ 7 } - \int { \frac { { x }^{ 6 } }{ 7 } dx }\) Integrate.
\(\frac { { x }^{ 7 }ln(9x) }{ 7 } - (\frac { { x }^{ 7 } }{ 49 } +C)\) For some constant 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 ) = $
Solution:
function_7 <- function(x)
{
1 / (6 * x)
}
integrate(function_7, lower = 1, upper = exp(6))## 1 with absolute error < 9.3e-05f(x) is a probability density function on the interval [1, e ^ 6] since the integral of f(x) on the interval is equal to 1.