Question

How do you use limits to evaluate `int(x^2+4x-2)dx` from [1,4]?

Answer 1

Here is a limit definition of the definite integral. (I don't know if it's the one you are using.)

.`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_1^4 (x^2+4x-2) dx`.

Find `Delta x`

For each `n`, we get

`Deltax = (b-a)/n = (4-1)/n = 3/n`

Find `x_i`

And `x_i = a+iDeltax = 1+i3/n = 1+(3i)/n`

Find `f(x_i)`

`f(x_i) = (x_i)^2 + 4(x_i) - 2 = (1+(3i)/n)^2 + 4(1+(3i)/n) - 2`

`= 1+(6i)/n+(9i^2)/n^2 +4+(12i)/n-2`

`= (9i^2)/n^2 + (18i)/n + 3`

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( (9i^2)/n^2 + (18i)/n + 3) 3/n`

`= sum_(i=1)^n( (27i^2)/n^3 + (54i)/n^2 + 9/n)`

`=sum_(i=1)^n (27i^2)/n^3 + sum_(i=1)^n(54i)/n^2 + sum_(i=1)^n 9/n`

`=27/n^3sum_(i=1)^n i^2 + 54/n^2 sum_(i=1)^n i + 9/n sum_(i=1)^n 1`

Evaluate the sums

`= 27/n^3((n(n+1)(2n+1))/6) + 45/n^2((n(n+1))/2) + 15/n(n)`

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

Rewrite before finding the limit

`sum_(i=1)^n f(x_i)Deltax = 27/n^3((n(n+1)(2n+1))/6) + 54/n^2((n(n+1))/2) + 9/n(n)`

`= 27/6((n(n+1)(2n+1))/n^3) + 54/2((n(n+1))/n^2) + 9`

`= 9/2((n(n+1)(2n+1))/n^3) + 27((n(n+1))/n^2) + 9`

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

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

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

To finish the calculation, we have

`int_1^4 (x^2+4x-2) dx= lim_(nrarroo) (9/2((n(n+1)(2n+1))/n^3) + 27((n(n+1))/n^2) + 9)`

`= 9/2(2) + 27(1) + 9`

`= 45`