\[Answer1\] \[∫4e^{−7x}dx\\ u=−7x \\ du=−7dx \\ dx=-1/7du\]

\[∫4e^{−7x}dx \\ = ∫4e^u(−1/7du)\\ =-4/7∫e^udu\\ =−4/7e^u+C\\=−4/7e^{−7x}+C\]

\[Answer2\]

library(pracma)
## Warning: package 'pracma' was built under R version 4.3.3
dN_dt = function(t){
  -3150 / t^4 - 220
  }

n_t = function(t) {
    int = integral(dN_dt, 1, t)
    n_1 = 6530
    n_t = int + n_1
    return(n_t)
}

n_t(8)
## [1] 3942.051

\[Answer3\] \[a=4.5\\ b=8.5\\f(x)=2x−9\\∫^{8.5}_{4.5} \ 2x−9dx=\\ [x^2−9x]^{8.5}_{4.5}\\=[8.5^2−9(8.5)]−[4.5^2−9(4.5)]\\ =−4.25−(−20.25)=16=Area\] \[Answer4\] \[y=x^2−2x−2\\ y=x+2\] \[x^2−2x−2=x+2\\x^2−3x−4=0\\(x−4)(x+1)=0\\x=4\\and\\x=−1\] \[∫^4_{-1}x+2−(x^2−2x−2)dx\\ =∫^4_{-1}x+2−x^2+2x+2dx\\=∫^4_{-1}−x^2+3x+4dx\\=−x^3/3+(3/2)x^2+4x|^4_{−1}\\=[−4^3/3+3/2(4)^2+4(4)]−[−(−1)^3/3+3/2(−1)-4]=[−64/3+24+16]−[1/3+3/2−4]=20.8333\]

\[Answer5\]

# demand
d = 110
# order cost
c = 8.25
# storage cost
s = 3.75

# lot size 
size = sqrt((2*d*c)/s)
print(size)
## [1] 22
# number of orders per year 
n = d/size
print(n)
## [1] 5

\[Answer6\] \[∫ln(9x)⋅x^6dx\\ =∫[ln(9)+ln(x)]x^6dx\\ =∫ln(9)x^6dx+∫ln(x)x^6dx \\first\ integral: ln(9)/7x^7\\ second\ integral\\ u=ln(x)\\du=1/xdx\\ dv=x^6dx\\ v=x^7/7 \\x^7ln(x)/7−∫x^7/7*1/xdx=\\x^7ln(x)/7−∫x^6/7dx=x^7ln(x)/7−x^7/49+C \\ Combine \ the\ two\ results: =x^7ln(9x)/7−x^7/49+C\] \[Answer7\] \[f(x)=1/6x\] \[∫^{e^6}_1=1/6\ ln(x)|^{e^6}_1\\=1/6\ ln(e^6)-1/6ln(1)\\=1-0=1\\The\ area\ under f(x) in \ the \ interval\ of\ [1,e^6]= 1\ means \ f(x)=1/6x\ is\ a\ probability\ density \ function \ on\ the\ interval \ [1,e^6] .\] ```