# (a) Use Definition 2 to find an expression for the area under the curve y= (x^3) from 0 to 1 as a limit. Then usethe following formula (1^3) + (2^3) + (3^3) +...... + (n^3) = [(n(n+1))/2]^2 to...

(a) Use Definition 2 to find an expression for the area under the curve y= (x^3) from 0 to 1 as a limit. Then use

the following formula

(1^3) + (2^3) + (3^3) +...... + (n^3) = [(n(n+1))/2]^2

to evaluate the limit.

### 1 Answer | Add Yours

You should create a partition of the interval [0,1] in n subintervals that have the following lengths such that:

`Delta x_i = (1-0)/n = 1/n`

`x_i = i*(1/n)`

You need to use limit definitionto evaluate the definite integral such that:

`int_0^1 x^3 dx = lim_(n->oo) sum_(i=1)^n f(x_i)*Delta x_i`

`int_0^1 x^3 dx = lim_(n->oo) sum_(i=1)^n (i^3/n^3)*(1/n` )

`int_0^1 x^3 dx = lim_(n->oo) 1/n^4sum_(i=1)^n i^3`

You should remember the formula that gives the summation `sum_(i=1)^n i^3` such that:

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

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

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

`int_0^1 x^3 dx = (1/4)lim_(n->oo) 1*(1 + 1/n)*(1 + 1/n) = 1/4`

**Hence, evaluating the area under the curve, using the limit definition of definite integral, yields `int_0^1 x^3 dx = 1/4.` **