Problem 1.

Use integration by substitution to solve the integral below.

\(\int { 4 } { e }^{ -7x }{ dx }\)

\(=\ 4\int { { e }^{ -7x }{ dx } }\)

Substitute and recall that integral of en is en:

\(u\ =\ -7x,\ du\ =\ -7dx\ or -\frac { 1 }{ 7 } du = dx\)

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

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

Resubstitute for u:

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


Problem 2.

Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \(\frac { dN }{ dt } =\quad -\frac { 3150 }{ { t }^{ 4 } } -220\) 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.

A. We integrate over the given velocity function:

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

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

\(=\ -3150\ \cdot \ -\frac { 1 }{ 3{ t }^{ 3 } } \ -\ 220\int { 1\ dt } \\\)

\(=\ -1050{ t }^{ -3 }\ -220t\ +\ 6530\)

\(\ 6530\ =\ -1050{ t }^{ -3 }\ -220t\ +\ C\)

\(\ 6530\ =\ -1270\ +\ C\)

\(\ C\ =\ 5260\)

\(\ N(t)\ =\ -1050{ t }^{ -3 }\ -220t\ +\ 5260\)


Problem 3:

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

A. Use Reimman Sums, although we have a strange interval, [4.5, 8.5]. Because our function bisects the height of each rectangle at midpoint, we will use the midpoint rule. This takes many steps.

\(\Delta x\ =\ \frac { b-a }{ n } =\frac { 8.4-4.5 }{ 4 } =1\)

\({ x }_{ i\ }=\ a+(i-1)\cdot \Delta x\ =\ 4.5+1\cdot (i-1)\ =\ i+3.5\)

\({ x }_{ i+1 }\ =\ i+1+3,5\ =\ i+4.5\)

\(\frac { { x }_{ i }{ +x }_{ i+1 } }{ 2 } =\frac { (i+3.5)\ +\ (i+4.5) }{ 2 } \ =\ i+4\)

\(\int _{ 4.5 }^{ 8.5 }{ 2x-9\ dx\ \approx } \ \sum _{ i=1 }^{ 4 }{ f( } \frac { { x }_{ i }{ +x }_{ i+1 } }{ 2 } )\ \Delta x\)

\(=\sum _{ i=1 }^{ 4 }{ f( } i+4)\ \Delta x\)

\(=\sum _{ i=1 }^{ 4 }{ f( } 2(i+4)-9)\ \Delta x\)

\(=\sum _{ i=1 }^{ 4 }{ f( } 2i-1)\ \Delta x\)

\(=\ \Delta x\ (2\cdot \sum _{ i=1 }^{ 4 }{ i\ } -\sum _{ i=1 }^{ 4 }{ 1 } )\)

\(=\ 1\cdot 2(\frac { 4\cdot 5 }{ 2 } )-4\cdot 1\)

\(=\ 16\)


Problem 4

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

y=x2-2x-2, y=x+2

First, plot it, then integrate and follow Theorem 52:

g <- makeFun(c^2-2*c-2 ~ c)
g2 <- makeFun(c+2 ~ c)

plotFun(g(x) ~ x,  x.lim = range(-2,5), lwd=5)

plotFun(g2(x) ~ x, add=TRUE, x.lim = range(-2,5), lwd=5, col='red')

ladd(panel.abline(v=0, col='gray50'))

ladd(panel.abline(h=0, col='gray50'))

ladd(panel.abline(v=-1, col='gray50'))

ladd(panel.abline(v=4, col='gray50'))

Area equals 20.8 units per R integration below.

integrandG <- function(x){x^2-2*x-2} 
G <- integrate(integrandG, lower = -1, upper = 4)

integrandG2 <- function(x){x+2} 
G2 <- integrate(integrandG2, lower = -1, upper = 4)

abs(G2$value-G$value)
## [1] 20.83333