Question

How do you use the limit process to find the area of the region between the graph `y=x^2+2` and the x-axis over the interval [0,1]?

Answer 1

Please see the explanation section below.

Explanation:

Here is a limit definition of the definite integral. (I'd guess it's the one you are using.) I will use what I think is somewhat standard 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.)

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

`int_0^1 (x^2+2) dx`.

Find `Delta x`

For each `n`, we get

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

Find `x_i`

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

Find `f(x_i)`

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

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

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

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

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

Evaluate the sums

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

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

Rewrite before finding the limit

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

`= 1/6(n/n*(n+1)/n*(2n+1)/n) +2`

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

`lim_(nrarroo) (n/n) = 1`

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

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

To finish the calculation, we have

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

`= 1/6((1)(1)(2))+2`

`= 1/3 + 2 = 7/3`.