3 ms·

Is there a good explanation of u substitution (or any other integral stuff that makes use of the denominator) using non-Liebnitz notation? Is it just that ∂1 is

• bikenaga 3y ago
  If you mean u-substitution in definite integrals, that is something called the change of variables theorem. I think you mean indefinite integrals or antiderivatives, and I think you're asking for an explanation with using the "du/dx" sort of notation. Here's a quick explanation which may be more than you wanted to know. :-) You have ∫ f(g(x)) g'(x) dx. [So it might be something like ∫ [cos(x^2)] (2 x) dx, where f(x) = cos x and g(x) = x^2.] You do the u-substitution u = g(x), so u'(x) = g'(x): ∫ f(g(x)) g'(x) dx = ∫ f(u) u'(x) dx You want to say that the second integral is ∫ f(u) du -- essentially (to go back to "fractional" derivative notation), you want to do (du/dx) dx = du to get rid of the x's. Why is this justified? Let F(u) = ∫ f(u) du: F is the antiderivative of f. By definition of the antiderivative, F'(u) = f(u). So by the Chain Rule, d/dx F(u) = F'(u) u'(x) = f(u) u'(x). But since F(u) = ∫ f(u) du differentiates to f(u) u'(x), by definition F(u) is the antiderivative of f(u) u'(x). But the antiderivative of f(u) u'(x) is ∫ f(u) u'(x) dx. So putting all this stuff together, ∫ f(g(x)) g'(x) dx = ∫ f(u) u'(x) dx = F(u) = ∫ f(u) du.