Almost afraid to ask, but shouldn't that integral be 1/9, not 1?

Alan7996

By weighted sum, are the weights determined by the interval h?

viceversa

If the number is finite, why not just add them together?

coolpotato

Just to clarify, the weighted sum that we use to approximate the integral is the sum over all sample points: (length of interval)*(sample value)?

Murrowow

Is this similar conceptually to doing Riemann sums or trapezoidal estimations of the area under a curve?

air-wreck

If polynomials are so easy to integrate, then could we just use a polynomial approximation (like the Taylor series) to a given curve to get an estimate? How does this compare to other numerical integration methods?

Almost afraid to ask, but shouldn't that integral be 1/9, not 1?

By weighted sum, are the weights determined by the interval h?

If the number is finite, why not just add them together?

Just to clarify, the weighted sum that we use to approximate the integral is the sum over all sample points: (length of interval)*(sample value)?

Is this similar conceptually to doing Riemann sums or trapezoidal estimations of the area under a curve?

If polynomials are so easy to integrate, then could we just use a polynomial approximation (like the Taylor series) to a given curve to get an estimate? How does this compare to other numerical integration methods?