What is the integral of x*sqrt (2x^2)?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to find the integral of x*sqrt (2x^2). This can be done using substitution. But I will use a simpler method.

f(x) = x*sqrt (2x^2)

=> f(x) = x*x*sqrt 2

=> f(x) = x^2*sqrt 2

Int[f(x) dx]

=> Int[x^2 * sqrt 2]

=> ((sqrt 2)/3)*x^3 + C

The integral of x*sqrt (2x^2) is ((sqrt 2)/3)*x^3 + C

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

This problem is a little bit tricky. Since it is not given the range of admissible values for x, we'll solve the integral admitting that interval of values for x is the real set number.

Since the domain of values for the function f(x) = x*sqrt(2x^2) is R, then:

sqrt x^2 = |x| and not x

We'll solve the integral considering 2 cases:

Case 1:

For x>0, |x| = x

Int x*sqrt(2x^2)dx = (sqrt2)*Int (x^2)dx = [(sqrt2)*x^3]/2 + C

Case 2:

For x<0. then |x| = -x

Int x*sqrt(2x^2)dx = (sqrt2)*Int - (x^2)dx = - [(sqrt2)*x^3]/2 + C

The indefinite integral of the given function is: [(sqrt2)*x^3]/2 + C, if x > 0, or - [(sqrt2)*x^3]/2 + C, if x < 0.

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