1
Integrate by substitution
\(\int 4e^{-7x}dx\)
\(u = -7x\)
\(\frac{du}{dx} = -7\)
\(du = -7dx\)
\(dx = -\frac{1}{7}du\)
\(\int e^u du = -\frac{4}{7}e^{-7x}\)
2
Find the function N(t)
Rate of change: \(\frac{dN}{dt} = -\frac{3150}{t^4} - 220\)
To find the formula the shows the relation between time and bacteria contamination rate, we need to find the anti-derivative of the rate of change in respect to time.
\(dN = (-\frac{3150}{t^4} - 220)dt\)
\(dN = (-3150t^{-4} - 220)dt\)
\(N = \frac{1050}{t^3}-220t + C\)
After the first day the contamination level is 6530 so we solve when t = 1.
\(6530 = 830 + C\)
\(C = 5700\)
\(N(t) = \frac{1050}{t^3} -220t + 5700\)
3
\(\int^4_0 2x-9\)
\(x^2 - 9x|^4_0\)
\(16\) since the area has to be positive
4
\(y = x^2 - 2x - 2, \ y = x + 2\)
To find where they intersect, set the equations equal to each other and find the roots.
\(x^2 - 2x - 2 = x + 2\)
\(x^2 - 3x - 4 = 0\)
\(x(x-3) = 4\)
Roots are -1, 4.
\(\int^4_1 x^2-3x-4\)
\(\frac{1}{3}x^3 - \frac{3}{2}x^2-4x\)
= 20.84
5
Average storage costs:
Since we don’t know the number of irons ordered we’ll divide the ‘x’ amount by 2 to find the average.
\(3.75 \times \frac{x}{2} = 1.875\) per iron
Ordering costs:
\(8.35 \times \frac{110}{x} = \frac{907.5}{x}\)
Minimum Cost:
C = \(1.875x + \frac{907.5}{x}\)
C’ =\(1.875 - \frac{907.5}{x^2}\)
\(1.875 = \frac{907.5}{x^2}\)
\(x^2 = \frac{907.5}{1.875}\)
x = 22 irons per lot
110/22 = 5 orders
6
\(\int ln(9x)x^6dx\)
Integrate by parts:
\(\int f dg = fg \int gdf\)
We’ll make
\(f = ln(9x)\)
\(dg = x^6\)
so,
\(df = \frac{1}{x} dx\)
\(g = \frac{1}{7}x^7\)
Therefore,
\(\frac{1}{7}x^7ln(9x) - \frac{1}{7}\int x^6dx\)
\(\frac{1}{7}\int x^6 = \frac{1}{7}\times \frac{1}{7} x^7\)
Answer is:
\(\frac{1}{7}x^7ln(9x) - \frac{1}{49}x^7 + C\)
7
The probablility over the distribution has to equal 1 so:
${e6}_1 f(x) = 1 $
Integrate
\(\int \frac{1}{6x} = \frac{1}{6}\int \frac{1}{x}dx\)
\(\frac{1}{6} log(x)\)
Calculate the interval
\(\frac{1}{6} log(e^6) = 1\)
\(\frac{1}{6} log(0) = 0\)
So, 1 - 0 = 1.
The function is a pdf over the integral.