calculus II

posted by .

Using integration by substitution.
find the exact value of

integral from [0,9/16]
sqrt(1 - sqrt(x))/(sqrt(x))

  • calculus II -

    Do a u substitution.

    u= 1- sqrt(x)
    du = -(1/(2*sqrt(x)))dx

    Change your limits by plugging them into the u equation.

    u= 1 - sqrt(0) = 1-0 = 1
    u= 1 - sqrt(9/16) = 1-(3/4) = 1/4

    Substitute the u values in for x.

    The new integral is -2*sqrt(u) du from [1,1/4]

    OR

    2*sqrt(u) du from [1/4,1]


    You integrate and get 2*(2/3)*u^(3/2) evaluated from [1/4,1]. Plug in 1, then plug in (1/4). Subtract these two values and you should get your answer.

    I got 7/6 or 1.166667

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus URGENT test tonight

    Integral of: __1__ (sqrt(x)+1)^2 dx The answer is: 2ln abs(1+sqrt(x)) + 2(1+sqrt(X))^-1 +c I have no clue why that is! Please help. I used substitution and made u= sqrt(x)+1 but i don't know what happened along the way! Your first …
  2. Math Help please!!

    Could someone show me how to solve these problems step by step.... I am confused on how to fully break this down to simpliest terms sqrt 3 * sqrt 15= sqrt 6 * sqrt 8 = sqrt 20 * sqrt 5 = since both terms are sqrt , you can combine …
  3. Calculus

    so we are doing integrals and I have this question on my assignment and I can't seem to get it, because we have the trig substituion rules, but the number isn't even so its not a perfect square and I just cant get it, so any help would …
  4. Calculus

    Graph the curve and find its exact length. x = e^t + e^-t, y = 5 - 2t, from 0 to 3 Length = Integral from 0 to 3 of: Sqrt[(dx/dt)^2 + (dy/dt)^2] dx/dt = e^t - e^-t, correct?
  5. Calculus

    Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 …
  6. Math/Calculus

    Solve the initial-value problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else?
  7. Calculus

    Find the volume of the solid whose base is the region in the xy-plane bounded by the given curves and whose cross-sections perpendicular to the x-axis are (a) squares, (b) semicircles, and (c) equilateral triangles. for y=x^2, x=0, …
  8. Calculus

    Evaluate the indefinite integral: 8x-x^2. I got this but I the homework system says its wrong:sqrt((-x-8)x)/(2*sqrt(x-8)*sqrt(x))*(((sqrt(x-8)*(x-4)*sqrt(x))-32*log(sqrt(x-8)+sqrt(x))
  9. Calculus

    Hi. In an integration solution, the integral of (1/(sqrt (8-u squared)) is written as arcsin(u/sqrt 8), but I don't see how they got it. When I did it I got (1/8)*(arcsin(u*sqrt8)). What I did was take sqrt8 common in the denominator …
  10. Calculus 2 (Differential Equation)

    How would you solve the following problem explicitly?

More Similar Questions