U=???7x dU=???7dx dx=dU/???7
4???eUdU/???7 4/???7???eUdU 4/???7eU+C 4/???7e???7x+C
dN/dt=3150/t4???220 dN=(3150/t4???220)dt N=???3150/t4dt??????220dt N=N0???3150/3t3???220t N=N0???31503/3t3???220t N(1)=N0???1050/13???220(1) N0=6530+1050+220 N0=7800 N=7800???1050/t^3???220t
3.Find the total area of the red rectangles in the figure below, where the equation of the line is f (x ) = 2x ??? 9.
f(x) = 2x-9.
A1=(5.5???4.5)???1=1 A2=(6.5???5.5)???3=3 A3=(7.5???6.5)???5=5 A4=(8.5???7.5)???7=7 Area=1+3+5+7=16
Area=???8.54.5(2x???9)dx
Area=(2???x^2/2???9x)|8.5 4.5 Area=(8.52???9???(8.5))???(4.52???9???(4.5)) Area=(72.25???76.5???20.25+40.5) Area=16
A=???4???1x+2dx??????4???1x2???2x???2dx A=1/2x2|4???1+2x|4???1???[1/3x3???x2???2x]|4???1 A=???[1/3x3???3/2x2???4x]|4???1 A=[3/2x2+4x???1/3x3]|4???1
((3/2)*4^2 +4*4 -(1/3)*4^3) - ((3/2)*(-1)^2 +4*(-1) -(1/3)*(-1)^3)
## [1] 20.83333
Let C be cost, r be the number of orders per year, and x be the number of at irons in an order. rx=110 so x=110/r, assume half an order is in storage at on average. Such that,
C=8.25r+3.75x/2 C=8.25r+206.25/r C’=8.25???206.2/5r^2 C’=0 r=sqrt(206.25/8.25) r=5 orders per year
6.Use integration by parts to solve the integral below.
ln(9x). x6dx ???ln(9x)???x6dx U=ln(9x) dU=1/xdx dV=x6dx V=1/7x7 ???UdV=UV??????VdU 1/7ln(9x)x7???1/7???x6dx 1/7x7[ln(9x)???1/7] ???
f (x) = 1/ 6x
F(x)=???e61f(x)dx=1 f(x)=1/6x F(x)=???e61 1/6xdx F(x)=1/6 ???e^6 1 1/xdx F(x)=1/6[ln(x)|e^6 1 F(x)=1/6[ln(e6)???ln(1)] F(x)=1/6[6???0]=1 The sum of the definite integral for the function f(x)=1/6x on the interval [0,e^6] is 1. Therefore, we can say that the function f(x) is a probability density function.