Solving a work problem using integrals? please help!

A pyramid has a square base with side length 756 ft and height 481 ft. The density of the limestone is about 150 lb/ft^3.
Estimate the total work done in building the pyramid.

I got a huge number: 1.347x10^11 ft-lb by integrating (9.8)(150)[(756/481)(481-x)]^2 dx from 0 to 481

if someone else could share his answer and how he got it...that'd be great. not sure if my thought process is correct...

First you have a unit problem. g=9.8m/s² is in SI units but feet and pounds are imperial. You can't mix them in the same equation as you have done.

Secondly, a foot pound is the amount of energy needed to raise a 1lb weight by 1 ft. The value of 'g' isn't used to find the work in ft lb. (Unlike SI units, where kg and newtons are related by 'g')

If you want the final answer in joules, all distances and masses must first be converted to metres and kg.

Consider a horizontal square 'slice' of the pyramid at height h. By proportion, each side of the square will measure:
756(1 - h/481)
= 756 - 1.5717h

The area of the slice = (756 - 1.5717h)²
= 571536 - 1188.2h + 2.470h²

If the slice has thickness dh, its volume = (571536 - 1188.2h + 2.470h²)dh
Since the density =150lb/ft³ the weight of each slice is
150 x (571536 - 1188.2h + 2.470h²)dh

The work done raising each slice a distance h is weight x h
= 150 x (571536h - 1188.2h² + 2.470h³)dh

The total work done raising all the slices is:
150 x Integral(571536h - 1188.2h² + 2.470h³)dh for h=0 to h=481
= 150 x [571536 x 481²/2 - 1188.2x481³/3 + 2.470x 481⁴/4]
= 150 x [6.61x10¹⁰ - 4.41x10¹⁰ + 3.31x10¹⁰]
= 8.27x10¹² ft lb
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However, I don't kow if all that arithmetic is correct.

Your equation is wrong, first thing is that you have 9.8 which is the metric term for 'g' in m/s^2, not in english units.