Untitled

1. Use integration by substitution to solve the integral below: ∫4e−7xdx

u=−7x

du=−7dx

dx=du−7

∫4eudu−7

4−7∫eudu

4−7eu+C

4−7e−7x+C

2. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of dNdt=−3150t4−220 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.

dNdt=−3150t4−220

dN=(−3150t4−220)dt

N=∫−3150t4−220dt

N=∫−3150t4dt−∫220dt

N=−31503t3−220t+C

N(1)=−31503(1)3−220(1)+C=6530

C=7800

N(t)=−31503(t)3−220(t)+7800

N(0)=−31503(0)3−220(0)+7800=7800

3. Find the total area of the red rectangles in the figure below, where the equation of the line is f ( x ) = 2x  9. ∫8.54.52x−9dx

=[x2−9x]|8.54.5

=(8.52−9(8.5))−(4.52−9(4.5))

=16

Area of red rectangles is 16

4. Find the area of the region bounded by the graphs of the given equations y=x2−2x−2, y=x+2

Graph for equation:

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 }


# zeros of f 
root <- polyroot(c(-4, -3, 1)) 
ifelse(Im(root) == 0, Re(root), root)

## [1] -1+0i  4-0i

Area: ∫4−1x+2dx−∫4−1x2−2x−2dx

=−[13x3−32x2−4x]|4−1

=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. Number of orders per year = n lot size = s Cost = c

ns = 110 s = 110 / n

Assume half of inventory keep in stocks:

c=8.25n+3752n

c=8.25n+206.25n

c′=8.25−206.25n2

c′=0

0=8.25−206.25n2

n=5

Number of order per year is 5

Lot size is 22 and inventory cost is 78.75.

6. Use integration by parts to solve the integral below ∫ln(9x)∗x6dx

Choose:

u=ln(9x), dvdx=x6

du=99xdx=1xdx

dv=x6dx

v=17x7

In equation:

∫udv=uv−∫vdu

=ln(9x)17x7−∫17x71xdx

==ln(9x)x77−x749−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)=16x

∫e61f(x)dx

=∫e6116xdx

=16∫e611xdx

=16ln(x)|e61

=16[ln(e6)−ln(1)]

=16[6−0]

=1