Question

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

Answer 1

For evaluation using the definition, please see the full explanation section below. (Skip to the end if you can use the Fundamental Theorem of Calculus.)

Explanation:

`int_1^2 (x^2+1)dx`.

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 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_1^2 (x^2+1)dx`.

For each `n`, we get

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

And `x_i = a+iDeltax = 1+i1/n =1+ i/n`

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

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

`sum_(i=1)^n f(x_1)Deltax = sum_(i=1)^n(i^2/n^2+(2i)/n+2) 1/n`

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

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

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

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

So,

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

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` and `lim_(nrarroo)[(n(n+1))/n^2]`

There are several ways to think about the first 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.

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`.

The second limit

The ideas above can be applied to get

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

Completing the integration

.`int_1^2(x^2+1)dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax`

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

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

`= 1/6 [2] + [1] + [2]`

`= 1/3+3 = 10/3`

Using the Fundamental Theorem of Calculus

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

`int_1^2(x^2+1) dx = (x^3/3+x)]_1^2`

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

`= 14/3-4/3=10/3`