Question

How to find the definite integral using the limit definition?

`int_1^4x^3-4dx`

Please explain in detail, I don't understand the process

Answer 1

Here is a limit definition of the definite integral. (I'd guess 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^3-4) 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)^3-4 = (1+(3i)/n)^3-4`

`= (1+9i/n+27i^2/n^2+27i^3/n^3) - 4`

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

`= sum_(i=1)^n( -9/n+27i/n^2+81i^2/n^3+81i^3/n^4)`

`=sum_(i=1)^n ( -9/n)+sum_(i=1)^n(27i/n^2)+sum_(i=1)^n(81i^2/n^3)+sum_(i=1)^n(81i^3/n^4)`

`=-9/nsum_(i=1)^n ( 1)+27/n^2sum_(i=1)^n(i)+81/n^3sum_(i=1)^n(i^2)+81/n^4sum_(i=1)^n(i^3)`

Evaluate the sums

`= -9/n(n) +27/n^2((n(n+1))/2) + 81/n^3((n(n+1)(2n+1))/6) +81/n^4( (n^2(n+1)^2)/4)`

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

Rewrite before finding the limit

`sum_(i=1)^n f(x_i)Deltax = -9/n(n) +27/n^2((n(n+1))/2) + 81/n^3((n(n+1)(2n+1))/6) +81/n^4( (n^2(n+1)^2)/4)`

`= -9 +27/2((n(n+1))/n^2) + 81/6((n(n+1)(2n+1))/n^3) +81/4( (n^2(n+1)^2)/n^4)`

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

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

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

`lim_(nrarroo) ((n^2(n+1)^2)/n^4) = 1`

To finish the calculation, we have

`int_0^1 x^2 dx = lim_(nrarroo) (-9 +27/2((n(n+1))/n^2) + 27/2((n(n+1)(2n+1))/n^3) +81/4( (n^2(n+1)^2)/n^4))`

`= -9+27/2(1) + 81/6(2)+81/4(1)`

`= -36/4+54/4+108/4+81/4 = 207/4`.