How do you integrate int2xe^x dx from 0 to 1? Calculus Techniques of Integration Integration by Parts 1 Answer Truong-Son N. ยท 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 2621 views around the world You can reuse this answer Creative Commons License