Question

need help with :definite integral from 0 to 1 x^3/ Square root of x^4+9

Answer 1

\[\int\limits_{0}^{1}{x^3 \over \sqrt{x^4+9}}dx\]Is this the correct equation?

Answer 2

yes, that is the one.

Answer 3

u-substitute for x^4 + 9

Answer 4

let u be x^4+9?

Answer 5

but how do you know which one to put as" u"?

Answer 6

u = x^4 + 9
du = 4x^3 dx
du / 4x^3 = dx\[\int\limits\limits_{0}^{1}{x^3 \over \sqrt{u}}{du \over 4x^3}\]\[\int\limits\limits_{0}^{1}{du \over 4\sqrt{u}}\]Can you finish it? I think you are actually supposed to re-evaluate the limits but I don't remember how to do that. It is not necessary if you replace u with what you substituted it for before you apply the limits

Answer 7

let me try replacing it