What is the definite integral of 2xe^4x with an upper limit or boundary of 1 and a lower limit or boundary of 1/4? Is it 3e^(4)/(8)?

Answer 1

Use integration by parts:

2∫xe^4xdx let u = x and dv = e^4xdx then du = dx and v = ∫dv = (1/4)e^4x

So, the original integral is equivalent to 2 [ (1/4)xe^4x - (1/4)∫e^4xdx ] evaluated from 1/4 to 1

= (1/2)xe^4x - (1/8)e^4x evaluated from 1/4 to 1

= (1/2)e⁴ - (1/8)e⁴ - (1/8)e + (1/8)e = (3/8)e⁴