As you can see all you have to do now is to integrate e^u with respect to du which is a simple and straightforward integration.Īt this stage we can now substitute u back in. ½∫ e^u du: This is just a bit of tidying up with ½ behind the integral. ∫ e^x^2 x dx ≡ ∫ e^u ½ du: All that I am doing here is replacing x^2 with u, and also replacing x dx with the expression found earlier. The goal here is to get an expression where x dx is on one side and all the unwanted terms on the other side, hence it is just a simple matter of transposition. The derivative of u is 2x which is a good indicator that substitution will work as there is a x dx in the problem. For beginners the main question seems to be, "How do you determine when to use the substitution method?" A rough rule is that if the derivative of u looks anything like the outside part x dx then substitution will have a good chance of working. Sometimes you need to change the order of. Hence, it is just for clarity for beginners so they can follow better this way. We conclude that the integral10ey1f(x,y)dxdy with integration order reversed is e11logxf(x,y)dydx. The purpose of moving it is to group x dx together as this is the part I will be substituting out. Using the u substitution method is the best way to solve it.Īll that I have done here is move x closer to dx and it is still the same expression. ∫ xe^x^2 dx: Integrating this is extremely simple and ideal for beginners. so we put u x2, and all is well: x 3e x2 dx x2× 1 2e x2 2x×12e x2 dx 12 xe u x2 v xex2 x2 x2 dx x2 x2 +c u 2x v 1 2 ex2 x2e.
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