1. Use integration by substitution to solve the integral below.

\[\int 4 e^{-7x}~dx\] This is the same as: \[-4/7\int -7 e^{-7x}~dx\]

If u = -7x, \[\frac{du}{dx} = -7 \] \[du = -7 dx \] And rearranging the problem we get: \[-4/7\int -7~dx e^{-7x}\] Which is the same as: \[-4/7\int du e^{u}\] And the integral of e^u is e^u so we can solve for the following: \[-4/7\int du e^{u} = -4/7e^{u} + C\] Or… \[= -4/7e^{-7x} + C\]

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.

\[dN = \int`\frac{3150}{t^4} - 220 dt\]
\[= \frac{-3150}{3t^3} - 220t + C \] We know at time, 1, the bacteria is 6530, which allows us to solve for C: \[6530= \frac{-3150}{3(1)^3} - 220(1) + C \] \[6530= -1050 -220 + C \] \[7800 = C\]

Plugging C back in, we see the function N(t) to estimate the level of contamination:

\[N(t) = \frac{-3150}{3t^3} - 220t + 7800 \]

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

Evaluating the integral at 8.5 minus 4.5 we get the total area is 16.

func <- function(x)(2*x-9)
integrate(func,4.5,8.5)
## 16 with absolute error < 1.8e-13

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

\(y = x^2 -2x -2\)
\(y = x + 2\)

Let’s first graph the functions. We see the graphs intersect at x = -1 and x = 4

eqn1 <- function(x){x^2 -2*x - 2}
eqn2 <- function(x){x + 2}
plot(eqn1(-5:5), type='l')
lines(eqn2(-5:5), type='l')

\[\int_{-1}^{4} x + 2 dx - \int_{-1}^{4} x^2 -2x - 2 dx\] Using the integrate function in R, we get an area of ~20.83

left_function <- function(x)(x+2)
right_function <- function(x)(x^2-2*x-2)
left <- integrate(left_function,-1,4)
right <- integrate(right_function,-1,4)
left$value - right$value
## [1] 20.83333

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.

Let n be the number of order and x be the number of flat irons. Cost = 3.75x + 8.25n

n = 110/x
x = 110/n

Substituting x for n we get: Cost = 3.75(110/n) + 8.25n Cost = 412.5/n + 8.25n

To minimize cost we take the deritiative and set the derivative equal to 0: Cost’ = -412.4/n^2 + 8.25 0 = -412.4/n^2 + 8.25 -8.25 = -412.4/n2 8.25n2 = 412.4 n^2 = 49.98788 n = ~7

x = 110/n x = 110/7 So they should do 7 orders with approximately 16 flat irons in each box.

6. Use integration by parts to solve the integral below

\[\int ln(9x)x^6dx\] \[ln(9x)\int x^6dx-\int \frac{1}{x} (\int x^6dx) dx\] \[ln(9x)(\frac {x^7}{7})-\int \frac{1}{x} (\frac {x^7}{7}) dx\] \[ln(9x)(\frac {x^7}{7})-\int (\frac {x^6}{7}) dx\] \[ln(9x)(\frac {x^7}{7})-(\frac {x^7}{49})\] \[(\frac {x^7}{7})(ln(9x)-\frac {1}{7})\]

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.

Yes, it’s the probability density function. \[ f(x) = \frac{1}{6x}\] \[\int_{1}^{e^6} \frac{1}{6x} dx \] \[= \frac{1}{6} (lnx)| [1,e^6] \] \[= \frac{1}{6}(0 - lne^6)\] \[= lne \] \[= 1\]