Question

How do you find the area between the given curve `y= x^2` and the x-axis given in the interval [0,1]?

Answer 1

The area is `1/3`

Explanation:

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.

.`int_a^b x^2 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^2dx`.

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_1)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]`

So,

`sum_(i=1)^n f(x_1)Deltax = 1/6 [(n(n+1)(2n+1))/n^3]`

The last thing to do is evaluate the limit as `nrarroo`.
I hope it is clear that this amounts to evaluating
`lim_(nrarroo)(n(n+1)(2n+1))/n^3`

There are several ways to think about this:

Limit of a Rational Expression

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.

OR

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

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

So the limit is, again `2`.

However we get it, we get

.`int_0^1 x^2 dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax`

`= lim_(nrarroo) sum_(i=1)^n((i^2)/n^2) 1/n`

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

`= 1/6 (2)`

`= 1/3`