Integrals question, help plz?

For geometric integration by area

You can take f(x)*dx from 0 to 5 and

Int(f(x)) = (f(5) - f(4))*(5-4) + (f(4) - f(3))*(4-3) + (f(3) - f(2))*(3-2) + (f(2) - f(1))*(2-1) + (f(1) - f(0))*(1-0)

f(x) = xi * ln(1+xi^2)

(f(5) - f(4))*(5-4) = (5 * ln(1+25) - 4 * ln(1+16))*1 = 4.958

(f(4) - f(3))*(4-3) = (4 * ln(1+16) - 3 * ln(1+9))*1 = 4.425

(f(3) - f(2))*(3-2) = (3 * ln(1+9) - 2 * ln(1+4))*1 = 3.689

(f(2) - f(1))*(2-1) = (2 * ln(1+4) - 1 * ln(1+1))*1 = 2.526

(f(1) - f(0))*(1-0) = (1 * ln(1+1) - 0 * ln(1+0))*1 = 0.693

Int(f(x)) = 4.958 + 4.425 + 3.689 + 2.536 + 0.693

Int(f(x)) = 16.30

This is just one way of looking at integration and makes the most sense to me.

The integral is best thought of at the area under a curve. The example above creates 5 trapezoids with heights of f(x=n) and f(x=n+1) and a width of 1. However to get accuracy, you decrease the width of the trapezoids.

You can find the integral or area under the curve by doing the following as well

int(f) = (f(5.0) - f(4.9))*(0.1) + (f(4.9) - f(4.8))*(0.1) + ... + (f(0.1) - f(0.0))*(0.1).

imperative of[(x+a million)/(x^2+a million)] =imperative of [x/(x^2+a million)]+imperative of[a million/(x^2+a million)] =I1+I2 for I1 enable x^2+a million=t then dieeerentiate it we get 2xdx=dt xdx=dt/2 I1=imperative of (a million/2t) I1=a million/2log(t) I1=a million/2log(x^2+a million) and I2=intergal of (a million/(x^2+a million)) I2=tan inverse(x) =>ans is: a million/2log(x^2+a million)+tan inverse(x)