How do you find the integral from 0 to 2 of #xe^(2x) dx#? Calculus Techniques of Integration Integration by Parts 1 Answer Equivirial Jun 6, 2015 #\int_0^2xe^{2x}dx=3/4e^4+1/4# Use integration by parts #\int u dv=uv-int vdu# Let #u=x, \implies du=dx# Let #dv=e^{2x}dx, \implies v=1/2e^{2x}# Substitute #v# and #u# into the top expression #\int_0^2xe^{2x}dx=[x/2e^{2x}]_0^2 -int_0^2 1/2e^{2x}dx# #\int_0^2xe^{2x}dx=(e^4-0)-[1/4e^{2x}]_0^2# #\int_0^2xe^{2x}dx=3/4e^4+1/4# 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 8775 views around the world You can reuse this answer Creative Commons License