Question

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

Answer 1

Answer:

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