The following formula for the sum of the cubes of the first n integers is proved. Use it to evaluate the area under the curve y=x^3 from 0 to 1 as a limit, 1^3+2^3+3^3 + ... + n^3=[n (n+1) / 2]^2

the Area is approximately 1

Step-by-step explanation:

Since

y=x^3

the area under a curve can be approximated to rectangles of width Δx and length y, then

Area = Sum of rectangles of width x and length = ∑ y * Δx

taking the width Δx=1 for each rectangle

Area = Sum of rectangles of width x and length = ∑ y = ∑ x^3

from x=0 to x=1, there is only one rectangle of width Δx=1, then

since ∑ x^3 from x=0 to x=n is n (n+1) / 2]^2

Area = ∑ x^3 from x=0 to x=1 = 1 (1+1) / 2]^2 = 1

Note:

- The actual value of the area is calculated through the integral

Area = ∫ y dx = ∫ x^3 dx from x=0 to x=1 = 1⁴/4 = 1/4