Calc

1

Solve by substitution

\(\int4e^{-7x}dx\)

Let:

\(z=-7x\)

Therefore:

\(dx=-\frac{1}{7}dz\)

\(4\int e^z \frac{-1}{7}dz\)

\(=\frac{-4}{7}\int e^zdz\)

\(= -\frac{4}{7}e^z+C=\)

\(= -\frac{4}{7}e^{-7x}+C\)

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.

\(\int\frac{dN}{dt}=\int\frac{-3150}{t^4}-220dt\)

\(= -\frac{12600}{t^3}-220t+C\)

Since \(N(1)=6530\)

\(N(t)=-\frac{12600}{t^3}-220t+6290\)

3

Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x) = 2x - 9\).

The height of the rectangles is 1, 3, 5, and 7

\(\int_{4.5}^{8.5} 2x-9 dx\)

\(= [x^2 - 9x]|_{4.5}^{8.5}\)

 area = (((8.5^2)-(9*(8.5)) - ((4.5^2)-(9*(4.5)))))
 area 
## [1] 16

4

Find the area of the region bounded by the graphs of the given equations. \(y = x 2 - 2x -2\)

\(y = x + 2\)

curve(x^2 -2*x-2, lwd = 2, xlim=c(-5, 5))
curve(x+2, lwd = 2, xlim=c(-5, 5), add = TRUE)

Intersection point: \(x^2 -2x-2 = x+2 x^2 -3x-4 = 0\)

f <- function(a){ a^2 - 3*a - 4 }

root <- polyroot(c(-4, -3, 1)) 
ifelse(Im(root) == 0, Re(root), root)
## [1] -1+0i  4-0i

Area =

\(\int_{-1}^{4}x+2 dx -\int_{-1}^{4}x^2 -2x-2 dx\)

\(=-[\frac{1}{3}x^3 - \frac{3}{2}x^2 -4x]|_{-1}^{4}\)

\(~21\)

5

A beauty supply store expects to sell 110 flat irons during the next year. It costs \(3.75\) dollars to store one flat iron for one year. There is a fixed cost of \(8.25\) dollars for each order.

Find the lot size and the number of orders per year that will minimize inventory costs.

fullfilment <-function(x){ 
  fix<-x*8.25 
  space<-0 
  i=0
  while(ceiling((110/x)-i*110/365)>0){ 
    space<-space+(3.75/365)*ceiling((110/x-i*110/365))
    i<-i+1
    if(i==10000){
      break 
    }
  }
  space<-space*x
  fix<-fix+space
  return(fix)
}
fullfilment (1)
## [1] 216.9144
fullfilment (2)
## [1] 122.0342
fullfilment (3)
## [1] 95.97945
fullfilment (4)
## [1] 87
fullfilment (5)
## [1] 84.91438
fullfilment (6)
## [1] 86.36301

There should be 5 orders of 22 flat irons each.

6

Use integration by parts to solve the integral below:

\(\int \ln(9x)x^6dx\)

Let: \(u=\ln(9x)=\ln(9)+\ln(x)\)

\(dv=x^6dx\)

\(v=\frac{x^7}{7}\)

\(du = \frac{1}{x} + ln(9)xdx\)

\(\int \ln(9x)x^6dx\) \(=(\frac{x^7}{7}\ln(9x))-\int\frac{x^7}{7}(\frac{1}{x}+\ln(9)x)dx\) \(=\frac{x^7}{7}\ln(9x))-\int\frac{x^6}{7}+\ln(9)\frac{x^8}{7}dx\) \(= (\frac{x^7}{7}\ln(9x))-\frac{x^7}{56}-\ln(9)\frac{x^9}{63}+C\)

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.

\(f(x)=\frac{1}{6x}\)

\(=\frac{1}{6}\int_1^{e^6}\frac{1}{x}dx\) \(= \frac{1}{6}(\ln({e}^{6})-\ln(1))\) \(=\frac{1}{6}(6\ln(e)-\ln(1))\) \(=\frac{1}{6}(6\times1-0)\)

\(=1\)