Question

How do you evaluate the definite integral by the limit definition given `int x^2+1dx` from [1,2]?

Answer 1

Please see the explanation section below.

Explanation:

Here is a limit definition of the definite integral.

.`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`.

Find `Delta x`

For each `n`, we get

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

Find `x_i`

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

Find `f(x_i)`

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

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

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

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

Evaluate the sums

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

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

Rewrite before finding the limit

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

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

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

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

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

To finish the calculation, we have

`int_0^4 (x^2+1) dx = lim_(nrarroo) ( 2+ ((n(n+1))/n^2) + 1/6 ((n(n+1)(2n+1))/n^3))`

`= 2+(1)+1/6(2) = 10/3`