\[ \int 4e^{-7x} \]
\[ = -\frac{1}{7} * 4e^{-7x} \]
\[ = -\frac{4}{7} * e^{-7x} \]
First we solve for N(t). I re-write the function as:
\[ \frac{dN}{dt} = 3,150 *t^{-4} - 220 \]
\[ N(t) = -\frac{1}{5}* 3,150 *t^{-5} - 220t + C \]
Now we need to solve for C
\[ 6530 = -630 -220+C \]
\[ C = 7,380 \]
Now we plug in C
\[ N(t) = -630 *t^{-5} - 220t + 7,380 \]
Here we need to evaluate the integral from 5 to 8.
\[ \int_{5}^{8} 2x-9dx \]
Evaluate from 5 to 8 \[ = x^2-9x \]
\[ (8^2-9(8)) - (5^2-9*(5)) \]
\[ (-8) - (-20) = 12 \]
First we want to find where the two functions intersect
\[ x^2 -2x -2 = x+2 \]
\[ x^2 -3x -4 = 0 \]
\[ (x-4)(x+1) = 0 \]
\[ x=-1, 4 \]
Now we need to solve the integral of the two functions
\[ \int_{-1}^{4} (x^2 -2x-2) - (x+2) dx \]
\[ \int_{-1}^{4} x^2 -3x -4 dx \]
\[ \frac{x^3}{3} - \frac{3x^2}{2} - 4x \]
\[ (\frac{4^3}{3} -\frac{4^2}{2} - 4*(4)) - (\frac{-1^3}{3} -\frac{-1^2}{2} - 4*(-1)) = 6.83 \]
Here x=number of orders and y=lot size. At any given time we have half the stock on hand. We have 2 functions: \[ x*y=110 , cost=8.25x + 3.75*(\frac{y}{2}) \]
solving for y gives us \[ cost = 8.25x + 3.75*\frac{110}{2x} \]
\[ cost = 8.25x + \frac{206.25}{x} \]
Solve the derivative: \[ cost = 8.25 - \frac{206.25}{x^2} \]
To minimize cost we set it equal to 0 and solve for x
\[ \frac{206.25}{x^2} = 8.25 \]
\[ \frac{206.25}{8.25} = x^2 \]
Solving gets x=5.
\[ \int ln(9x)*x^6dx \]
\[ u=ln(9x), u'=\frac{1}{x}, v'=x^6,v=\frac{1}{7}x^7 \]
\[ uv - \int u'vdx \]
\[ \frac{ln(9x)x^7}{7} - \int \frac{1}{x}\frac{x^7}{7} \]
\[ \frac{ln(9x)x^7}{7} - \int \frac{x^6}{7} \]
\[ \frac{ln(9x)x^7}{7} - \frac{x^7}{49} \]
\[ \frac{x^7}{7}(ln(9x) -\frac{1}{7}) \]
\[ \int_{1}^{e^6} \frac{1}{6x} \]
\[ = \frac{1}{6}ln(x) \]
\[ = \frac{1}{6}ln(e^6) - \frac{1}{6}ln(1) = 1-0 = 1 \]
That means f(x) is a probability function because the sum under the curve over a large span of x equals 1