Question

How do you evaluate `int f(x)=x^2 + 1` for [0,3]?

Answer 1

See the explanation for `int_0^3(x^2+1)dx` using the definition of definite integral.

Explanation:

I think you are asking to find

`int_0^3(x^2+1)dx`.

(If I've misunderstood your question, I am sorry.)

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 the definition.

.`int_0^3(x^2+1)dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax`.

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

For each `n`, we get

`Deltax = (b-a)/n = (3-0)/n = 3/n`

And `x_i = a+iDeltax = 0+i3/n = (3i)/n`

`f(x_i) = x_i^2+1 = ((3i)/n)^2+1 = (9i^2)/n^2+1`

`sum_(i=1)^n f(x_1)Deltax = sum_(i=1)^n((9i^2)/n^2+1)3/n`

`= sum_(i=1)^n((27i^2)/n^3+3/n)`

`= 27/n^3 sum_(i=1)^n i^2 + 3/nsum_(i=1)^n1)`

`= 27/n^3[(n(n+1)(2n+1))/6] + 3/n (n)`

So,

`sum_(i=1)^n f(x_1)Deltax = 9/2[(n(n+1)(2n+1))/n^3] + 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_0^3(x^2+1)dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax`

`= lim_(nrarroo) sum_(i=1)^n((9i^2)/n^2+1)3/n`

`= lim_(nrarroo) [9/2[(n(n+1)(2n+1))/n^3] + 3]`

`= 9/2(2)+3`

`= 12`

Answer 2

If you have the Fundamental Theorem of Calculus available the see the explanation below.

Explanation:

`int_0^3(x^2+1)dx = [x^3/3+x]_0^3`

`= [3^3/3+3] - 0^3/3+0]`

`= 9+3`

`= 12`