Question

How do you use the limit process to find the area of the region between the graph `y=16-x^2` and the x-axis over the interval [1,3]?

Answer 1

Here is a limit definition of the definite integral. (Others are possible.)

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

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_1^3 (16-x^2) dx`.

Find `Delta x`

For each `n`, we get

`Deltax = (b-a)/n = (3-1)/n = 2/n`

Find `x_i`

And `x_i = a+iDeltax = 1+i2/n = 1+(2i)/n`

Find `f(x_i)`

`f(x_i) = 16-(x_i)^2 = 16-(1+(2i)/n)^2`

`= 16-(1+(4i)/n+(4i^2)/n^2)`

`= 15 -(4i)/n - (4i^2)/n^2`

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

`sum_(i=1)^n f(x_i)Deltax = sum_(i=1)^n( 15 -(4i)/n - (4i^2)/n^2) 2/n`

`= sum_(i=1)^n( 30/n -(8i)/n^2 - (8i^2)/n^3)`

`=sum_(i=1)^n ( 30/n) - sum_(i=1)^n((8i)/n^2) - sum_(i=1)^n((8i^2)/n^3)`

`=30 /nsum_(i=1)^n ( 1)-8/n^2sum_(i=1)^n(i)-8/n^3sum_(i=1)^n(i^2)`

Evaluate the sums

`= 30/n(n) -8/n^2((n(n+1))/2) - 8/n^3((n(n+1)(2n+1))/6)`

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

Rewrite before finding the limit

`sum_(i=1)^n f(x_i)Deltax = 30/n(n) - 8/n^2((n(n+1))/2) - 8/n^3((n(n+1)(2n+1))/6)`

`= 30 - 4((n(n+1))/n^2) - 4/3((n(n+1)(2n+1))/n^3)`

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

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

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

To finish the calculation, we have

`int_0^1 x^2 dx = lim_(nrarroo) (30 - 4((n(n+1))/n^2) - 4/3((n(n+1)(2n+1))/n^3)`

`= 30 - 4(1) - 4/3(2)`

`= 90/3 - 12/3 - 8/3 = 70/3`.