Question

The integral from 0 to 1 of x^4r dx = 1/(4r+1)
I need to know how this works to solve another problem.

Answer 1

If \(r\neq-\dfrac{1}{4}\), then the power rule for integration gives
\[\int_0^1x^{4r}\,\mathrm{d}x=\bigg[\frac{x^{4r+1}}{4r+1}\bigg]_{x=0}^{x=1}=\frac{1^{4r+1}}{4r+1}-\frac{0^{4r+1}}{4r+1}=\frac{1}{4r+1}\]

Answer 2

Thank you, that helps@HolsterEmission