1

Use integration by substitution to solve the integral below:

$\int4e^{-7x}dx$

Solution:

We will use substitution $u=e^x$. $u'=e^x$.

$\int4e^{-7x}dx=\int4u^{-8}du=\frac{-4u^{-7}}{7}+c=\frac{-4e^{-7x}}{7}+c$

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:

$N(t)=\int-3150/t^4-220dt=3150/(3t^3)-220t+c$

$N(1)=6530=3150/3-220+c$, or $c=5260$

$N(t)=3150/(3t^3)-220t+5260$

6530-3150/3-220

## [1] 5260

3

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

Solution:

$\int_{4.5}^{8.5} 2x-9dx=x^2-9x=8.5^2-8.5*9-4.5^2+4.5*9=16$

8.5^2-8.5*9-4.5^2+4.5*9

## [1] 16

4

Find the area of the region bounded by the graphs of the given equations.

$y=x^2-2x-2$

$y=x+2$

Solution:

Let us first plot our functions.

$x^2-2x-2=0$

Points were the first function crosses x axis.

$x_1=(2+\sqrt{2^2+4*2})/2=2.732$

$x_2=(2-\sqrt{2^2+4*2})/2=-0.732$

## [1] 2.732051

## [1] -0.7320508

$x^2-2x-2=x+2$

$x^2-3x-4=0$

Points were two functions cross each other.

$x_1=(3+\sqrt{3^2+4*4})/2=4$

$x_2=(3-\sqrt{3^2+4*4})/2=-1$

Areas under the curves are:

$\int_{-1}^{-0.732} x^2-2x-2dx=(1/3)x^3-x^2-2x=0.131$

$\int_{-0.732}^{2.732} x^2-2x-2dx=(1/3)x^3-x^2-2x=6.928$

$\int_{2.732}^{4} x^2-2x-2dx=(1/3)x^3-x^2-2x=3.464$

$\int_{-1}^4 x+2dx=x^2/2+2x=17.5$

## [1] 0.1307683

## [1] -6.928203

## [1] 3.464102

## [1] 17.5

## [1] 20.83333

## [1] 17.5

## [1] 0.123

## [1] -6.915518

## [1] 3.06068

## [1] 21.23184

Answer: 20.8333

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 the number of orders per year that will minimize inventory costs.

Solution:

$f(x)=8.25x+(110/x)3.75$, where x is number of orders, 0<x<111

$f'(x)=8.25-(110/x^2)3.75=0$

$x=\sqrt{110*3.75/8.75}=6.9$

(110*3.75/8.75)^.5

## [1] 6.866066

test<-c()

for (i in (1:110))
{test<-c(test,(8.25*i+110*3.75/i))
}

match(min(test),test)

## [1] 7

Answer: 7

6

Use integration by parts to solve the integral below

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

Solution:

$u=x^7$; $u'=7x^6$; $v=ln(9x)/7$

$\int ln(9x)x^6dx=\int vdu=vu-\int udv=x^7 ln(9x)/7-\int x^7dln(9x)/7=x^7ln(9x)/7-\int x^6/7dx=x^7ln(9x)/7-x^7/49+c$

Answer: $x^7ln(9x)/7-x^7/49+c$

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.

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

Solution:

$\int_1^{e^6} 1/6xdx=lnx/6=6/6=1$, as integral of our function is equal to 1 on interval $[1,e^6]$, $f(x)$ is a probability density function.

Data 605. Assignment for Week 13. 04/30/2019.

Mikhail Groysman

April 30, 2019

1

2

3

4

5

6

7