Problem 1
Use integration by substitution to solve the integral below.
\[\int 4{ e }^{ -7x }dx\]
Solution
Let u = -7x, then du = -7dx. In terms of dx, dx = \(\frac{-1}{7}\)
Substitute u and du into the integral:
= \(\int 4{ e }^{ -7x }dx = \int 4{ e }^{u}(\frac{du}{-7})\)
= \(-\frac{4}{7}\int 4{ e }^{u}du\)
= \(-\frac{4}{7}*e^{u}+C\)
Substitute back in terms of x for final solution:
= \(-\frac{4}{7}*e^{-7x}+C\) (where C is the constant of integration)
Problem 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:
\(\int dN\) = -3150 \(\int\frac{1}{t^{4}-220}dt\)
substitute u=\(t^2\), then du/dt=2t and dt=du/(2t)
\(\int dN\) = -3150 \(\int\frac{1}{u^{2}-220}\frac{du}{2t}\)
N = -3150 \(\int\frac{1}{u^{2}-220}du\)
N = -3150 \(\frac{1575}{u-\sqrt{220}} + C\)
Note that u=\(t^2\)
N = -\(\frac{1575}{t^{2}-\sqrt{220}} + C\)
For level of contamination after 1 day (t=1) was 6530 bacteria per cubic centimeter
6530 = -\(\frac{1575}{1-\sqrt{220}} + C\)
C = 6530 + \(\frac{1575}{1-\sqrt{220}}\)
The function to estimate the level of contamination is:
N(t) = 6530 + \(\frac{1575}{1-\sqrt{220}} - \frac{1575}{t^{2}-\sqrt{220}}\)
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
Solution:
Area = 1+3+5+7 = 16sq units
Problem 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’s find the points of intersection between the two curves. Set y in both equations equal to each other:
==> \(x^{2} − 2x − 2 = x + 2\)
= \(x^{2}−2x-x-2−2 = 0\)
= \(x^{2}−3x-4 = 0\)
= \((x-4)(x+1) = 0\)
= \(x = -1, 4\)
Now, we can find the area of the region by integrating the difference between the upper curve (x+2) and the lower curve (x^{2}-2x−2) over the interval of (−1,4):
==> \(\int_{-1}^{4} [(x^{2}-2x-2)-(x+2)]dx\)
= \(\int_{-1}^{4}(x^{2}-3x-4)dx\)
= \([\frac{x^{3}}{3} - \frac{3x^{2}}{2} - 4x]_{-1}^{4}\)
= \((\frac{64}{3} - \frac{3*16}{2} - 16) - (\frac{-1}{3} - \frac{3}{2} +4)\)
= -20.833
The area of the region bounded by the graphs of the given equations is 20.83333.
Problem 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:
Using the Economic Order Quantity (EOQ) = \(\sqrt{\frac{2*Demand * Order cost\ per\ order}{Holding\ cost\ per\ unit}}\)
==> \(\sqrt{\frac{2*110 * 8.25}{3.75}}\)
≈ 22
To find the number of orders per year, we can use the formula:
Number of Orders= \(\frac{Demand}{EOQ}\)
==> 110/22
= 5
Therefore, to minimize inventory costs, the store should order 22 flat irons at a time, and there will be 5 orders per year.
Problem 6
Use integration by parts to solve the integral below. \[\int{ln(9x) . x^{6} dx}\]
Let u= ln(9x), then \(\frac{du}{dx}=\frac{1}{x}\)
Also, let \(\frac{dv}{dx}=x^6\) Therefore, v = \(\int{x^6 dx} = \frac{x^7}{7}\)
= \(\int{ln(9x).x^6dx} = ln(9x).\frac{x^7}{7} - \int{\frac{x^7}{7}\times\frac{dx}{x}}\)
= \(ln(9x) \times \frac{x^7}{7} - \int{\frac{x^6}{7}dx}\)
= \(ln(9x) \times \frac{x^7}{7} - \frac{x^7}{49} + k\)
= \(\frac{x^7}{49}(7ln(9x) - 1) + k\)
Problem 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.
\[f( x ) = \frac{1}{6x}\]
Soliution:
==> \(\int_1^{e^6}\frac{1}{6x} dx = \frac{1}{6} ln(x)|_1^{e^6}\)
= \(\frac{1}{6} ln(e^6) - \frac{1}{6} ln(1)\)
= \(\frac{1}{6} \times 6 - \frac{1}{6} \times 0\)
= 1