Timeline for Why do no students know to change the limits of integration when doing substitutions?
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| Feb 12, 2018 at 3:21 | comment | added | Mark McClure | Sorry. :) For me, the definite integral is much more fundamental than the indefinite. I suppose that's why I think it's silly not to change bounds. | |
| Feb 12, 2018 at 3:10 | comment | added | Mike Pierce | It's killing me that you're right, because my intuition is still screaming otherwise. I think my intuition based only on how I originally learned to evaluate integrals via substitution is a bit off then, because I'm struggling to assimilate the statement and conditions of that identity into how I currently think about "doing $u$-substitution". | |
| Feb 12, 2018 at 2:30 | comment | added | Mark McClure | @MikePierce Thanks for the fun example! In your translation, though, I suspect you replaced $x$ with $\sqrt{u}$, which is valid for only half of the interval. No such mistake is necessary for my integral. Regardless, in the identity $$\int_{\varphi(a)}^{\varphi(b)}f(x)dx = \int_a^b f(\varphi(t))\varphi'(t) dt,$$ there is no assumption of monotonicity on $\varphi$. | |
| Feb 12, 2018 at 2:03 | comment | added | Mike Pierce | Making that substitution $u$ for $\sin(x)$ doesn't seem valid since $\sin(x)$ isn't injective on $(0,\pi)$. I don't think you can just "fold" the domain of integration like that. Here's a silly example: $$\text{Letting } u = x^2,\quad \int_{-1}^1 x^2 \;\mathrm{d}x \;=\; 2\!\int_1^1 \sqrt{u} \;\mathrm{d}u \;=\;0\,.$$ It only works out to be true for $\int_0^{\pi} \sin\left(e^{\sin(x)}\right)\cos(x)\,\mathrm{d}x$ because of the symmetry of the function about $\pi/2$. Doing the substitution $u = x-\pi/2$, the integrand becomes and odd function. | |
| Feb 12, 2018 at 1:43 | comment | added | Mark McClure | @MikePierce Well, you need to expose them to similar examples. When I teach integration, I spend plenty of time on geometry. They know that the integral of an odd function over an interval centered on the origin is zero. They know that the integral of a constant is the length of the interval times the constant. They most certainly know that $$\int_0^0\sin(e^u)du=0,$$ which is what that integral transforms into. When I teach Calc II, more than half would get that right. But then, I make them change bounds of integration from day one. | |
| Feb 12, 2018 at 1:22 | comment | added | Mike Pierce | How do your students respond to that second example of evaluating $\int_0^{\pi} \sin\left(e^{\sin(x)}\right)\cos(x)\,\mathrm{d}x$? Or maybe more importantly how do you expect them to respond? The students I've TAed (intro to calculus) wouldn't have a clue how to handle that correctly. | |
| Feb 11, 2018 at 22:18 | history | edited | Mark McClure | CC BY-SA 3.0 |
added 237 characters in body
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| Feb 11, 2018 at 20:01 | history | answered | Mark McClure | CC BY-SA 3.0 |