Question

How do you use limits to find the area between the curve `y=x^2` and the x axis from [0,5]?

Answer 1

Please see below.

Explanation:

Here is a limit definition of the area (actually for the definite integral). I will use what I think is frequently used notation in US textbooks.

.`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.)
We'll do one small step at a time.

`int_0^5 x^2 dx`.

Find `Delta x`

For each `n`, we get

`Deltax = (b-a)/n = (5-0)/n = 5/n`

Find `x_i`

And `x_i = a+iDeltax = 0+i5/n = (5i)/n`

Find `f(x_i)`

`f(x_i) = {:x_i:}^2 = ((5i)/n)^2`

`= (25i^2)/n^2`

Find and simplify `sum_(i=1)^n f(x_i)Deltax` in order to evaluate the sum.

`sum_(i=1)^n f(x_i)Deltat = sum_(i=1)^n ((25i^2)/n^2) 5/n`

`= sum_(i=1)^n( (125i^2)/n^3)`

`= 125/n^2 sum_(i=1)^n(i^2)`

Evaluate the sum

`= 125/n^3((n(n+1)(2n+1))/6)`

(We used a summation formula in the previous step.)

Rewrite before finding the limit

`sum_(i=1)^n f(x_i)Deltax = 125/6((n(n+1)(2n+1))/n^3)`

`= 125/6((n(n+1)(2n+1))/n^3)`

Now we need to evaluate the limit as `nrarroo`.

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

`= (1)(1)(2) = 2`

To finish the calculation, we have

`int_0^5 x^2dx = lim_(nrarroo)(125/6((n(n+1)(2n+1))/n^3))`

`= 125/6(2) = 125/3`