Calculus

posted by .

Evaluate
integral of x times square root of (2x-1)dx

  • Calculus -

    x (2x-1)^.5 dx

    u = x
    dv = (2x-1)^.5 dx

    du = dx

    v = (1/3)(2x-1)^1.5

    so
    (x/3)(2x-1)^(1.5) -(1/3) (2x-1)^1.5 dx

    (x/3)(2x-1)^1.5-(1/3)(1/5)(2x-1)^2.5

    (1/3)[ x(2x-1)^1.5 - (1/5)(2x-1)^2.5 ]

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus

    Find the exact area of the region enclosed by the square root of (x) + the square root of (y) is = 1; x = 0 and y = 0. I moved the first equation around to get: y = [1- the square root of (x) ] ^2 Unfortunately, this gives me bounds …
  2. Calculus

    evaluate the integral of dx/square root of 9-8x-x^2
  3. Calculus

    Evaluate the integral. The integral from 0 to the square root of 2 over 2 of the function 8/sq. rt.(1-t^2)
  4. Calculus

    Evaluate the integral from 0 to 24 of the square root of 9+3x dx.
  5. Calculus

    Evaluate integral of x times the square root of (x-4) dx I got u-(8u^(3/2) over 3) Can anyone check this for me?
  6. Calculus check

    The functions f and g are given by f(x)=sqrt(x^3) and g(x)=16-2x. Let R be the region bounded by the x-axis and the graphs of f and g. A. Find the area of R. B. The region R from x=0 to x=4 is rotated about the line x=4. Write, but …
  7. Calculus

    Which of the following integrals cannot be evaluated using a simple substitution?
  8. Calculus

    Which of the following integrals cannot be evaluated using a simple substitution?
  9. Calculus

    What is the best substitution to make to evaluate the integral of the quotient of cosine of 2 times x and the square root of the quantity 5 minus 2 times the sine of 2 times x, dx?

More Similar Questions