Post a New Question

Cal

posted by .

USE INTEGRATION BY PARTS TO FIND EACH INTEGRAL
ƪxe^2x dx

  • Cal -

    Please see your next post as there is the same problem.

    Look at Related Questions below for they look similar?

    Sra

  • Cal -

    let u = x
    du/dx = 1
    du = dx

    let dv = e^(2x)dx
    dv/dx = e^(2x)
    v = (1/2)e^(2x)

    formula:
    [int] u dv = uv - [int] v du
    [int] x e^(2x) dx = (x)(1/2)e^(2x) - [int] (1/2)e^(2x) dx
    = (1/2)xe^(2x) - (1/4)e^(2x)

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. integration by parts

    s- integral s ln (2x+1)dx ? = ln(2x+1)x - s x d( ln (2x+1)) = ln(2x+1)x- s x [(2x+1)'/ (2x+1)] dx = ln(2x+1)x- s x [(2)/ (2x+1)] ?
  2. calc

    how do you start this problem: integral of xe^(-2x) There are two ways: 1) Integration by parts. 2) Differentiation w.r.t. a suitably chosen parameter. Lets do 1) first. This is the "standard method", but it is often more tedious than …
  3. calc

    evaluate the integral: y lny dy i know it's integration by parts but i get confused once you have to do it the second time Leibnitz rule (a.k.a. product rule): d(fg) = f dg + g df y lny dy = d[y^2/2 ln(y)] - y/2 dy ----> Integral …
  4. Math - Indefinite Integration

    Use integration by parts to solve: I = f (x^3)(e^(x^2)) dx (Let "f" represent the integral) Let u = (e^(x^2)), dx = (1) / (2x(e^(x^2))) du Let dv = x^3, v = (x^4) / (4) I = (1/4)(x^4)(e^(x^2)) - (1/4) f (x^3) (2(e^(x^2))) du ... and …
  5. CALCULUS

    USE INTEGRATION BY PARTS TO FIND EACH INTEGRAL ƪxe^2x dx
  6. CALCULUS

    USE INTEGRATION BY PARTS TO FIND EACH INTEGRAL ƪx^5 Lnx dx
  7. calculus

    USE INTEGRATION BY PARTS TO FIND EACH INTEGRAL ƪãx lnx dx
  8. cal

    USE INTEGRATION BY PARTS TO FIND EACH INTEGRAL ƪsquare rootx lnx dx
  9. Maths - Integration by parts

    Use the partial ractions to find The integral (3x^2-x+2)/(x-1)(x^2+1) dx
  10. Integration by Parts

    integral from 0 to 2pi of isin(t)e^(it)dt. I know my answer should be -pi. **I pull i out because it is a constant. My work: let u=e^(it) du=ie^(it)dt dv=sin(t) v=-cos(t) i integral sin(t)e^(it)dt= -e^(it)cos(t)+i*integral cost(t)e^(it)dt …

More Similar Questions

Post a New Question