How do you integrate int2xe^x dx from 0 to 1?

1 Answer
Apr 5, 2015

By parts:

int_a^b 2xe^xdx = uv - int_a^bvdu

Let u = 2x and v = e^x.
du = 2dx

int_0^1 2xe^x dx = (2x*e^x) - int_0^1e^x*2 dx
= [(2x*e^x) - 2int_0^1e^x dx] eval (0->1)
= [2(1)e^(1) - 2(0)e^(0)] - 2(e^1 - e^0)
= [2e] - 2e + 2
= 2