Question

How do you use the limit process to find the area of the region between the graph `y=2x-x^3` and the x-axis over the interval [0,1]?

Answer 1

See 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_0^1 (2x-x^3) dx`.

Find `Delta x`

For each `n`, we get

`Deltax = (b-a)/n = (1-0)/n = 1/n`

Find `x_i`

And `x_i = a+iDeltax = 0+i1/n = i/n`

Find `f(x_i)`

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

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

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

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

Evaluate the sums

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

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

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

To finish the calculation, we have

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

`= (1)-1/4(1) = 3/4`