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

Question

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

1 Answer

Impact of this question

2446 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-05T21:03:38

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#

Science

Math

Humanities

... and beyond

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

1 Answer

Related questions

Impact of this question