Question 1:

Write a program to compute the derivative of f(x) = \({ x }^{ 3 }+\quad { 2x }^{ 2 }\) at any value of x. Your function should take in a value of x and return back an approximation to the derivative of f(x) evaluated at that value. You should not use the analytical form of the derivative to compute it. Instead, you should compute this approximation using limits.

options(warn = -1)
suppressMessages(library(mosaic))




diff <- function(x, h=1e-6){

    funcNeg <- (x-h)^3 + 2*(x-h)^2  
    funcPos <- (x+h)^3 + 2*(x+h)^2
    func     <- x^3 + 2*x^2
    upper_diff <- (funcPos - func)/h
    lower_diff <- (funcNeg - func)/-h
    diff_X <- (lower_diff + upper_diff)/2
    return(diff_X)
}

diff(2)
## [1] 20

Checking

funct <- D((x^3 + 2*x^2)~x)

funct(2)
## [1] 20

Question 2:

Now, write a program to compute the area under the curve for the function \({ 3x }^{ 2 }+\quad { 4x }\) in the range x = [1; 3]. You should first split the range into many small intervals using some really small \(\Delta\)x value (say 1e-6) and then compute the approximation to the area under the curve.

AUC <- function(){
   
    a <- 1
    b <- 3
    h <- 1e-6
    intervals <- (b-a)/h 
    area <- 0
    for(i in 1:intervals){
        area <- area + (3*a^2 + 4*a)*h 
        a <- a + h
    }
    return(area)
}

AUC()
## [1] 41.99998

Checking

integrate2 <- function(x)(3*x^2 + 4*x)
integrate(integrate2, lower=1, upper=3)
## 42 with absolute error < 4.7e-13

Question a:

  • Use integration by parts to solve for \(\int { sin(x)cos(x)dx }\)

Solution:

we shall make us of \(\int { f(g(x).g\prime (x)dx=\int { f(u)du,\quad u=g(x) } }\)

\(u=sin(x),\quad du=cos(x)dx\\ \frac { du }{ dx } \quad =\quad cos(x),\quad du=cos(x)dx,\quad dx=\frac { 1 }{ cos(x) } du\)

\(\int { ucos(x).\frac { 1 }{ cosx } } \quad =\quad \int { udu }\)

Recall that,

\(\int { { x }^{ a } } dx\quad =\quad \frac { { x }^{ a+1 } }{ a+1 } \quad =\quad \frac { { u }^{ 1+1 } }{ 1+1 } ,\quad since\quad u\quad =\quad sin(x)\)

\(\frac { { sin(x) }^{ 2 } }{ 2 } +C\)

Question b:

  • Use integration by parts to solve for \(\int { { x }^{ 2 }{ e }^{ x } } dx\)

Solution:

Recall from Intergration by part, that;

\(\int { { uv }^{ \prime } } =\quad uv\quad -\int { { vu }^{ \prime } }\)

where,

\(u\quad ={ x }^{ 2 },\quad { u }^{ \prime }=\quad 2x,\quad { v }^{ \prime }={ e }^{ x },\quad v={ e }^{ x }\)

\(=\quad { x }^{ 2 }{ e }^{ x }\quad -\quad \int { 2x{ e }^{ x } } dx\)

note that \(\quad \int { 2x{ e }^{ x } } dx\) = \(2( { e }^{ x }x\quad -\quad { e }^{ x })\)

Therefore,

\(\int { { x }^{ 2 }{ e }^{ x } } dx\) = \(\quad { x }^{ 2 }{ e }^{ x }\quad\) - \(2( { e }^{ x }x\quad -\quad { e }^{ x })\) + C

Question C:

  • What is \(\frac { d }{ dx } (xcos(x))\) ?

solution:

Recall from Product Rule,

\((f.g)\prime =f\prime .g\quad +f.g\prime \\ d(\frac { d }{ dx } (x)cos(x)x\quad +\quad \frac { d }{ dx } (cos(x)(x)(x)),\quad where,\\ =\quad -sin(x)(x)\quad +\quad 1.cos(x)\\ =-xsin(x)\quad +\quad cos(x)\)

Question d:

  • What is \(\frac { d({ e }^{ { x }^{ 4 } }) }{ dx }\) ?

Solution;

\(\frac { df(u) }{ dx } =\frac { df }{ du } .\frac { du }{ dx } ,\quad let\quad u={ x }^{ 4 }\\ =\quad \frac { { d(e }^{ u }) }{ du } \frac { { d(x }^{ 4 }) }{ dx } ,\quad \frac { d }{ du } { (e }^{ u })\quad ={ e }^{ u },\quad \frac { d }{ du } { (x }^{ 4 })\quad ={ 4x }^{ 3 }\\ =\quad { e }^{ u }.{ 4x }^{ 3 },\quad but\quad u={ x }^{ 4 }\\ =\quad { e }^{ { x }^{ 4 } }{ 4x }^{ 3 }\)