Calculus

what is the definite integral of
1)
Upper boundry 4
Lower boundry 0
Problem: x/ square root of 1+2x) dx
2)
Upper boundry 1/2
Lower boundry 0
Problem: sin^-1x / square root of 1-x^2) dx
3)
upper boundry 4
lower boundry 1
Problem: e^square root of x/ square root of x) dx
4)
upper boundry e^4
lower boundry e
Problem: dx/x square root of LN of x)
5) upper boundry pi/2
lower boundry -pi/2
Problem: (x^2 sin x)/ (1+x^6) dx

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asked by Tara
  1. 1)
    Upper boundry 4
    Lower boundry 0
    Problem: x/ square root of 1+2x) dx


    x/sqrt(1+2x) = 1/2 2x/sqrt(1+2x) =

    1/2 (2x+1-1)/sqrt(1+2x) =

    1/2 [sqrt(1+2x) - 1/sqrt(1+2x)]

    Integral of this is:

    1/6(1+2x)^(3/2) - 1/2 sqrt(1+2x)

    Insert upper and lower boundaries and subtract.


    arcsin(x)/sqrt(1-x^2)

    1/sqrt(1-x^2) is the derivative of arcsin(x), so the integral is

    1/2 arcsin^2(x)


    Integral of exp(sqrt(x))/sqrt(x) is

    2 exp(sqrt(x))


    Integral of 1/[x sqrt(ln(x))] is

    2 sqrt[ln(x)]

    Problem 5) The function you are integating is odd (i.e. the function changes sign when you change the sign of x), therefore the integral from
    -pi/2 to zero will cancel against the integral from zero to pi/2. So, the integral from -pi/2 to pi/2 is zero.

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