Question

tion(4): Indefinite and Definite Integration 4.1) Find the antiderivative: 3eis +3x5+ 4.2) The marginal revenue from the sale of x units estimated to be of a particular commodity is R(x) 50+ 3.5xe-001x dollars per unit (a) Find R(x) assuming R(O) 0 (bj What revenue should be expected from the sale of 1,000 units? Answers"***SHOW ALL WORK FOR FULL CREDIT

Answer 1

4.1)

we have

$\int_{1}^{3} 3e^{-0.15x}+3x^\frac{1}{2}-5+\frac{1}{3x}dx=\left [ \frac{3e^{-0.15x}}{-0.15}+\frac{3x^\frac{3}{2}}{\frac{3}{2}}-5x+\frac{1}{3}ln(x) \right ]_{1}^{3}$

$\int_{1}^{3} 3e^{-0.15x}+3x^\frac{1}{2}-5+\frac{1}{3x}dx=\left [ -20e^{-0.15x}+2x^\frac{3}{2}-5x+\frac{1}{3}ln(x) \right ]_{1}^{3}$

$\int_{1}^{3} 3e^{-0.15x}+3x^\frac{1}{2}-5+\frac{1}{3x}dx=\left [ -20e^{-0.15(3)}+2(3)^\frac{3}{2}-5(3)+\frac{1}{3}ln(3) \right ]-\left [ -20e^{-0.15(1)}+2(1)^\frac{3}{2}-5(1)+\frac{1}{3}ln(1) \right ]$

$\int_{1}^{3} 3e^{-0.15x}+3x^\frac{1}{2}-5+\frac{1}{3x}dx=-20e^{-0.45}+2(3)^\frac{3}{2}-15+\frac{1}{3}ln(3) +20e^{-0.15}-2+5-0$

$\int_{1}^{3} 3e^{-0.15x}+3x^\frac{1}{2}-5+\frac{1}{3x}dx=-20e^{-0.45}+2(3)^\frac{3}{2}+\frac{1}{3}ln(3) +20e^{-0.15}-12$

$\int_{1}^{3} 3e^{-0.15x}+3x^\frac{1}{2}-5+\frac{1}{3x}dx=3.22011$

4.2)

we have

$R'(x)=50+3.5xe^{-0.01x^2}$

integrate both side,

$\int R'(x)dx=\int 50+3.5xe^{-0.01x^2}dx$

$R(x)=50x+\int 3.5xe^{-0.01x^2}dx$ ................1)

now,

$\int 3.5xe^{-0.01x^2}dx=3.5\int xe^{-0.01x^2}dx$

let assume that,

$u=0.01x^2$

$du=0.02xdx$

$\frac{du}{0.02}=xdx$

$50du=xdx$

we can say that,

$\int 3.5xe^{-0.01x^2}dx=3.5\int e^{-u}50du$

$\int 3.5xe^{-0.01x^2}dx=175\int e^{-u}du$

$\int 3.5xe^{-0.01x^2}dx=-175e^{-u}+C$

substitute back u= 0.01x²,

$\int 3.5xe^{-0.01x^2}dx=-175e^{-0.01x^2}+C$

put this value in equation 1),

$R(x)=50x-175e^{-0.01x^2}+C$ ...............2)

we have R(0) = 0,

$R(0)=50(0)-175e^{-0.01(0)^2}+C$

$0=0-175(1)+C$

$C=175$

put C = 175 in equation 2),

$R(x)=50x-175e^{-0.01x^2}+175$

put x = 1000,

$R(1000)=50(1000)-175e^{-0.01(1000)^2}+175$

$R(1000)=50000)-175e^{-10000}+175$

$R(1000)=50175-\frac{175}{e^{10000}}$

$R(1000)=50175-0$

$R(1000)=50175$