Question

How do you evaluate the definite integral `int x^2 dx` from `[0,1]`?

Answer 1

I am assuming that you do not yet have the Fundamental Theorem of Calculus available to evaluate this, but that you need to evaluate it from a definition.

Explanation:

.`int_a^b f(x) dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax`.

Where, for each positive integer `n`, we let `Deltax = (b-a)/n`

And for `i=1,2,3, . . . ,n`, we let `x_i = a+iDeltax`. (These `x_i` are the right endpoints of the subintervals.)

I prefer to do this type of problem one small step at a time.

`int_0^1 x^2 dx`.

For each `n`, we get

`Deltax = (b-a)/n = (1-0)/n = 1/n`

And `x_i = a+iDeltax = 0+i1/n = i/n`

`f(x_i) = (x_i)^2 = (i/n)^2`

`= i^2/n^2`

`sum_(i=1)^n f(x_i)Deltax = sum_(i=1)^n(i^2/n^2) 1/n`

`= sum_(i=1)^n(i^2/n^3)`

`= 1/n^3 sum_(i=1)^n i^2`

`= 1/n^3[(n(n+1)(2n+1))/6]`

(We used summation formulas for the sums in the previous step.)

So,

`sum_(i=1)^n f(x_i)Deltax = 1/n^3[(n(n+1)(2n+1))/6]`

`= 1/6[(n(n+1)(2n+1))/n^3]`

The last thing to do is evaluate the limit as `nrarroo`.

There are a couple of ways to think about this limit :

The numerator can be expanded to a plynomial with leading term `2n^3`, so the limit as `nrarroo` is `2`.

OR

`(n(n+1)(2n+1))/n^3 = (n/n)((n+1)/n)((2n+1)/n)`

The limit at infinity is `(1)(1)(2)=2` as a product of rational expressions.

To finish the calculuation, we have

`int_0^1 x^2 dx = lim_(nrarroo)1/6[(n(n+1)(2n+1))/n^3] = 1/6[2] =1/3.`

Using the Fundamental Theorem of Calculus

Find an antiderivative of `x^2` (call it `F(x)`) and evaluate `F(1)-F(0)`

`int_1^2(x^2) dx = {: (x^3/3)]_0^1`

`= 1/3`