Question

How do you evaluate the definite integral by the limit definition given `int 3x^2+2dx` from [-1,2]?

Answer 1

See 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_-1^2 (3x^2+2) dx`.

Find `Delta x`

For each `n`, we get

`Deltax = (b-a)/n = (2-(-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+2 = 3(-1+(3i)/n)^2+2`

`= 3(1-(6i)/n +(9i^2)/n^2)+2`

`= 5-(18i)/n+(27i^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)Deltax = sum_(i=1)^n( 5-(18i)/n+(27i^2)/n^2) 3/n`

`= sum_(i=1)^n( 15/n - (54i)/n^2+(81i^2)/n^3)`

`=sum_(i=1)^n (15/n) - sum_(i=1)^n ((54i)/n^2) + sum_(i=1)^n((81i^2)/n^3)`

`=15/n sum_(i=1)^n (1) - 54/n^2 sum_(i=1)^n(i)+81/n^3 sum_(i=1)^n(i^2)`

Evaluate the sums

# = 15/n (n) -54/n^2((n(n+1))/2) +81/n^3 ((n(n+1)(2n+1))/6)#

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

Rewrite before finding the limit

#sum_(i=1)^n f(x_i)Deltax = 15/n (n) -54/n^2((n(n+1))/2) +81/n^3 ((n(n+1)(2n+1))/6)#

# = 15 -27((n(n+1))/n^2) +27/2 ((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`

`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_0^1 (x^2 +2) dx = lim_(nrarroo) (15 -27((n(n+1))/n^2) +27/2 ((n(n+1)(2n+1))/n^3))#

`= 15-27(1)+27/2(2)`

`= 15`.