Question

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

Answer 1

`int_-1^1x^3dx = 0`

Explanation:

We have,
`int_-1^1x^3dx = [x^4/4]_-1^1`

`= [(-1)^4/4 - (1)^4/4]`
`= [1/4 - 1/4]`
`= 0`

NB; If you want the AREA bounded by the curve, between x=-1 to x=1 and the x-axis, then the answer is
A=`2int_0^1x^3dx = 1/2` (by symmetry)

Answer 2

Please see the explanation section below.

Explanation:

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_-1^1 x^3 dx`.

Find `Delta x`

For each `n`, we get

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

Find `x_i`

And `x_i = a+iDeltax = -1+i2/n = -1+(2i)/n`

Find `f(x_i)`

`f(x_i) = (x_i)^3 = (-1+(2i)/n)^3 = -1+6i/n-12i^2/n^2+8i^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( -1+6i/n-12i^2/n^2+8i^3/n^3) 2/n`

`= sum_(i=1)^n( -1/n+6i/n^2-12i^2/n^3+8i^3/n^4)`

`=sum_(i=1)^n ( -1/n)+sum_(i=1)^n(6i/n^2)-sum_(i=1)^n(12i^2/n^3)+sum_(i=1)^n(8i^3/n^4)`

`=-1/nsum_(i=1)^n ( 1)+6/n^2sum_(i=1)^n(i)-12/n^3sum_(i=1)^n(i^2)+8/n^4sum_(i=1)^n(i^3)`

Evaluate the sums

`= -1/n(n) +6/n^2((n(n+1))/2) - 12/n^3((n(n+1)(2n+1))/6) +8/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 = -1/n(n) +6/n^2((n(n+1))/2) - 12/n^3((n(n+1)(2n+1))/6) +8/n^4( (n^2(n+1)^2)/4`

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

`= -1+3(1)-2(2)+2(1) = -1+3-4+2 =0`