Use integration by substitution to solve the integral below.
\[ \int4e^{-7x}dx \]
When solving integrals with substitution, it is useful to identify when some form of the derivative of one component of the function being integrated also exists within the function. In this case the derivative of the power to which \(e\) is being raised, \(-7x\), is the constant \(-7\), making it relatively easy to work with in this context (because \(4\) is itself a scalar). I can set \(-7x\) equal to \(u\) and reframe my integral in terms of \(du\).
Because \(u=-7x\), the differential \(du\) is equal to \(-7dx\). \(dx\) therefore equals \(-\frac{1}{7}du\). Plugging these values back into our integral, we get:
\[ \int4e^{u}(-\frac{1}{7}du) \]
Pulling out the scalars, we get:
\[ -\frac{4}{7}\int e^udu \] And because the antiderivative of \(e^u\) is \(e^u\), we are left with (accounting for an arbitrary scalar inherent to indefinite integrals):
\[ -\frac{4}{7}e^u + C \]
Plugging our \(x\) term back in for \(u\), we have:
\[ -\frac{4}{7}e^{-7x} + C \]
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.
Given a formula tracking the rate of change of the bacteria (our derivative), we can find the level of bacteria \(N\) at any time \(t\) by taking the antiderivative or integral of that derivative. While the result will be indefinite, we can also use the problem’s point-in-time value to make it definite.
\[ N(t) = \int-\frac{3150}{t^4}-220\:dt \]
The sum rule of integration tells us we can break up an integral of sums into the sum of integrals.
\[ \int-\frac{3150}{t^4}\;dt+ \int-220\:dt \]
Re-writing the first integral function as a negative exponent makes the application of the reverse power rule more intuitive.
\[ \int-3150t^{-4}\;dt+ \int-220\:dt \]
Pulling out the scalars:
\[ -3150\int t^{-4}\;dt-220\int1\:dt \]
Integrating using the reverse power rule:
$$ N(t) =-3150 * -t^{-3} - 220t + C \
N(t) = -220t+C $$ The above represents our generalized equation \(N(t)\). Now, we can use the additional information provided in the question to find \(C\).
\[ N(1) = 6530 \\ \\ 6530 = \frac{1050}{(1)^3}-220(1)+C \\ 6530 = 1050-220+C 6530 = 830 + C C = 5700 \] Therefore, the equation for the amount of bacteria after any number of days \(t\) is given by:
\[ N(t) = \frac{1050}{t^3}-220t+5700 \] 3. Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x)=2x-9\)
We can use the midpoint rule to solve this problem. There are 4 rectangles with the same width, which we can find by dividing the total length in question by 4. Because we appear to be approximating the area under the function from \(x=4.5\) to \(x=8.5\) and there are 4 rectangles, each must be \((8.5-4.5) / 4 = 1\) unit wide. The heights are given by the function \(f(x)\) at each of the values \(5\), \(6\), \(7\) and \(8\).
\[ f(5) = 2(5) - 9 = 10-9 = 1 \\ f(6) = 2(6) - 9 = 12-9 = 3 \\ f(7) = 2(7) - 9 = 14-9 = 5 \\ f(8) = 2(8) - 9 = 16-9 = 7 \]
Since each rectangle is \(1\) unit wide, the area of each is simply its height in units squared. Therefore, the total area of the rectangles is given by:
\[ 1+3+5+7 = 16\:units^2 \]
Find the area of the region bounded by the graphs of the given equations. \[ y = x^2 - 2x - 2 \\ y = x + 2 \] The area bounded by two functions is given by the difference of their integrals over a defined integral. To find that integral, we must first determine at what \(x\) values the functions intersect. We can do this by setting the functions equal to each other and solving:
\[ x^2 - 2x - 2 = x + 2 \\ x^2 - 3x - 4 = 0 \\ (x-4)(x+1) = 0 \\ x = -1, 4 \] In order to determine which integral to subtract from which, we can quickly check which function is greater in the interval between where they intersect. Choosing \(x=0\), we can see that \((0)+2=2\) is greater than \((0)^2 -2(0) - 2= -2\), so the integral of the quadratic function will be subtracted from that of the linear. Because of the additive properties of integrals, we can conduct this all in one integral:
\[ A = \int_{-1}^4 (x+2) \:- (x^2-2x-2)\:dx \\ = \int_{-1}^4 x+2-x^2+2x+2\:dx \\ = \int_{-1}^4 -x^2+3x+4\:dx \\ -\frac{1}{3}x^3 +\frac{3}{2}x^2+4x \Big|_{-1}^4 \] Now we can solve for \(F(4)-F(-1)\) where \(F\) is the antiderivative function we found through integration.
\[ F(4) = -\frac{1}{3}(4)^3 +\frac{3}{2}(4)^2+4(4) \\ = -\frac{1}{3}(64) + \frac{3}{2}(16) + 16 \\ = -21\frac{1}{3} + 40 \\ = 18\frac{2}{3} \] Now, \(F(-1)\):
\[ F(-1) = -\frac{1}{3}(-1)^3 +\frac{3}{2}(-1)^2+4(-1) \\ = \frac{1}{3}+\frac{3}{2}-4 = -2\frac{1}{6} \] And finally:
\[ F(4)-F(-1) \\ 18\frac{2}{3} + 2\frac{1}{6} \\ \approx 20.83 \]
\[ A = \int_{-1}^4x+2\:dx \:- \int_{-1}^4x^2-2x-2\:dx \\ = \frac{1}{2}x^2+2x \Big|_{-1}^4 - (\frac{1}{3}x^3 -x^2-2x)\Big|_{-1}^4 \\ \] For readability, I’ll break this up and start with the first half:
\[ \frac{1}{2}x^2+2x \Big|_{-1}^4 \\ \frac{1}{2}(4)^2+2(4) - (\frac{1}{2}(-1)^2+2(-1)) \\ 16 - (-\frac{3}{2}) \\ \frac{35}{2} \] Now the second half:
\[ \frac{1}{3}x^3 -x^2-2x)\Big|_{-1}^4 \\ \frac{1}{3}(4)^3 -(4)^2-2(4) - (\frac{1}{3}(-1)^3 -(-1)^2-2(-1)) \\ \frac{64}{3} -16-8-(-\frac{1}{3}-1+2) \\ \frac{64}{3}-24-\frac{2}{3} \\ -\frac{4}{3} \]
And finally:
\[ \frac{35}{2} - (-\frac{4}{3}) \\ \frac{35}{2} + \frac{4}{3} \\ \approx18.83 \]
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.
This problem is solvable using the Economic Order Quantity (EOQ) Model, which is as follows:
\[ EOQ = \sqrt{\frac{2DS}{H}} \] Where \(D\) is the yearly demand, \(S\) is the cost per order, and \(H\) is how much it costs to hold a unit for a year.
In this context, our annual demand \(D=110\), the cost-per-order \(S=8.75\), and the yearly storage cost for one unit \(H=3.75\). Given those values, we can solve for the \(EOQ\):
\[ EOQ = \sqrt{\frac{2(110)(8.25)}{3.75}} \\ = \sqrt{\frac{1815}{3.75}} \\ = \sqrt{484} \\ = 22 \] The \(EOQ\) of \(22\) means that \(22\) flat irons is the best amount per order to minimize costs, AKA the lot size.
In order to find the number of orders per year that minimize inventory costs, we simply divide the annual demand \(D\) by the \(EOQ\).
\[ \frac{D}{EOQ} \\ \frac{110}{22} \\ 5 \] Therefore, the best number of orders per year is \(5\).