How do you integrate #int2xe^x dx# from 0 to 1? Calculus Techniques of Integration Integration by Parts 1 Answer Truong-Son N. · Tom 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# Answer link Related questions How do I find the integral #int(x*ln(x))dx# ? How do I find the integral #int(cos(x)/e^x)dx# ? How do I find the integral #int(x*cos(5x))dx# ? How do I find the integral #int(x*e^-x)dx# ? How do I find the integral #int(x^2*sin(pix))dx# ? How do I find the integral #intln(2x+1)dx# ? How do I find the integral #intsin^-1(x)dx# ? How do I find the integral #intarctan(4x)dx# ? How do I find the integral #intx^5*ln(x)dx# ? How do I find the integral #intx*2^xdx# ? See all questions in Integration by Parts Impact of this question 2381 views around the world You can reuse this answer Creative Commons License