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  1. Use integration by substitution to solve the integral below. \[ \begin{aligned} \int 4 e^{-7 x} d x & \\ \int 4 e^{-7 x} d x &=\int \frac{-7 \times 4}{-7} e^{-7 x} d x \\ &=\int \frac{-4}{7} e^{u} d u \\ &=\frac{-4}{7} e^{u}+\text { constant } \\ &=-\frac{4}{7} e^{-7 x}+\text { constant } \end{aligned} \]

\[ =-\frac{4}{7} e^{-7 x}+\text { constant } \] 2. Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \(\frac{d N}{d t}=-\frac{3150}{t^{4}}-220\) 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. \[ \begin{aligned} N(t) &=\frac{1050}{t^{3}}-220 t+C \\ N(1) &=6530 \\ \frac{1050}{1^{3}}-220 \times 1+C &=6530 \\ C &=6530-1050+220 \\ C &=5700 \end{aligned} \]

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

#Find area in-build function
f3 = function(x) {2*x -9}

#Find the difference between areas under the curve
area3 <- integrate(f3, 4.5, 8.5)$value
area3 <- round(as.numeric(area3))
print(area3)
## [1] 16
  1. Find the area of the region bounded by the graphs of the given equations. Enter your answer below. \[ y=x^{2}-2 x-2, y=x+2 \]
#Find area in-build function
f1 = function(x) {x + 2}
f2 = function(x) {x^2 -2*x -2}

#Find the difference between areas under the curve
area1 <- integrate(f1, -1, 4)
area2 <- integrate(f2, -1, 4)
area <- round((area1$value - area2$value),4)
print(area)
## [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. \[ \begin{aligned} f^{\prime}(x) &=1.875-\frac{907.5}{x^{2}} \\ f^{\prime}(x) &=0 \\ 907.5 &=0 \\ x^{2} &=\frac{907.5}{x^{2}} \\ 1.875 &=\frac{907.5}{1.875 x^{2}}=\\ x^{2}=& \frac{907.5}{907.5} \\ x=& \sqrt{\frac{1.875}{}} \\ x=& \sqrt{484} \\ x=& 22 \end{aligned} \]

  1. Use integration by parts to solve the integral below. \[ \begin{aligned} \int_{\ln (9 x) \times x^{6} d x} &=\frac{1}{7} x^{7} \times \ln (9 x)-\int \frac{1}{7} x^{7} \times \frac{1}{x} d x \\ &=\frac{1}{7} x^{7} \times \ln (9 x)-\int \frac{1}{7} x^{6} d x \\ &=\frac{7}{49} x^{7} \times \ln (9 x)-\frac{1}{49} x^{7}+\text { constant } \\ &=\frac{1}{49} x^{7}(7 \ln (9 x)-1)+\text { constant } \end{aligned} \]
  2. Determine whether \(f(x)\) is a probability density function on the interval \(\left[1, e^{6}\right]\). If not, determine the value of the definite integral. \[ \begin{aligned} f(x) &=\frac{1}{6 x} \\ \int_{1}^{e^{6}} \frac{1}{6 x} d x &=\left.\frac{1}{6} \ln (x)\right|_{1} ^{e^{6}} \\ &=\frac{1}{6} \ln \left(e^{6}\right)-\frac{1}{6} \ln (1) \\ &=\frac{1}{6} \times 6-\frac{1}{6} \times 0 \\ &=1 \end{aligned} \]