Calculus

Numerical Integration and Differentiation

Write Programs

1. Write a program to compute the derivative of \(f(x) = x^3 + 2*x^2\) 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.

#Own function with limit

diff = function(f,value) {
  
ex = substitute_q(D(f,'x'),list(x=value))

if(is.nan(eval(ex))){ 
  
  #change value if the limit is infinite or NAN
  return(do.call("diff",list(f,0.000001)))
}

return(eval(ex))
  
}

#polynomial expression - change if required
f = expression((x^3 + 2*x^2))

#call the function
diff(f,0)

## [1] 0

diff(f,4)

## [1] 64

#Manual derivative
diff_manual <- function(x) {
    f = x^3 + 2*x^2
    h = 0.000001
    ex = ((x^3 +(3*h*x^2)+(3*x*h^2)+ h^3+ 2*(x^2 + (2*x*h)+h^2) - (x^3 + 2*x^2)) / h)
    return(ex)
}

#call the function
diff_manual(0)

## [1] 2.000001e-06

diff_manual(4)

## [1] 64.00001

2. Now, write a program to compute the area under the curve for the function \(3*x^2 + 4*x\) 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.

#Find area in-build function
integral = function(x) {3*x^2 + 4*x}

#Area under the curve is
integrate(integral,1,3)

## 42 with absolute error < 4.7e-13

#Integral with 

sum=0

for(x in c(seq(1, 3, 0.000001))){
  #apply small delta value to the integrated function
  sum = sum+ (((x^3) + 2*x^2) - (((x-0.000001)^3) + 2*(x-0.000001)^2))
}

#Area under the range[1,3]
sum

## [1] 42.00001

fn_plot = data.frame(seqq = c(seq(-3, 3, 0.001)))

fn_plot$curve = 3*fn_plot$seqq^2 + 4*fn_plot$seqq

fn_plot$curve_integral = fn_plot$seqq^3 + 2*fn_plot$seqq^2

ggplot(fn_plot,aes(x=seqq,y=curve,z=curve_integral)) + geom_line(aes(x=seqq,y=curve)) 

ggplot(fn_plot,aes(x=seqq,y=curve,z=curve_integral))+ geom_line(aes(x=seqq,y=curve_integral))

Solve Analytically

3. Please solve these problems analytically (i.e. by working out the math) and submit your answers.

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

#Find area in-build function
integral = function(x) {sin(x) * cos(x)}

#Area under the curve is
integrate(integral,1,3)

## -0.3440793 with absolute error < 7e-15

\[\begin{aligned} f(x) = \int sin(x)cos(x)dx\\ u = sin(x)\\ du = cos(x)\\ F(x) = \int u*du\\ F(x) = (u^2)/2\\ F(x) = sin(x)^2/2 + C\\ \end{aligned}\]

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

\[\begin{aligned} f(x) = \int x^2 e^x dx\\ u = x^2\\ du = 2x dx\\ v = e^x\\ dv = e^x dx\\ Formula: uv -\int vdu\\ \int x^2 e^x dx = x^2 e^x - \int 2x e^x dx\\ Subsitute \space again \space for \int 2x e^x dx\\ u = 2x\\ du = 2 dx\\ v = e^x\\ dv = e^x dx\\ \int 2x e^x dx = 2x e^x - \int 2 e^x dx\\ Final \space integral \space output\\ F(x) = x^2 e^x - 2xe^x - 2e^x \\ F(x) = e^x(x^2-2x-2)\end{aligned}\]

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

\[\begin{aligned} f(x) = \frac{d}{dx}(xcos(x))\\ f^*(x) = x *-sin(x) + cos(x)\\ f^*(x) = cos(x) - xsin(x)\end{aligned}\]

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

\[\begin{aligned} f(x) = \frac{d}{dx}*(e^{x^4})\\ f^*(x) = 4x^3e^{x^4}\end{aligned}\]