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Could someone give me a geometric interpretation of:

a) Integration by Parts

b) Integration by Substitution

Thanks!

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  • $\begingroup$ do you know geometric interpretation of integration? $\endgroup$ – dato datuashvili Jan 5 '14 at 12:50
  • $\begingroup$ clearly it depend on situation,substitution is used to simplify integral or convert difficult integration into easier one,the same for integration by part,generally we know that integral represent area on some bounded interval,therefore all these methods have same conceptual idea $\endgroup$ – dato datuashvili Jan 5 '14 at 13:01
  • $\begingroup$ I'm not really sure Dave. $\endgroup$ – dfg Jan 24 '14 at 19:23
  • $\begingroup$ With Integration by Parts do you mean en.wikipedia.org/wiki/Numerical_integration where the length x1- x2 is cut into parts. (PS This is not the normal use "Integration by Parts") For Integration by substitution see en.wikipedia.org/wiki/Integration_by_substitution if this doesn't help then please improve your question (add some examples you don't understand. $\endgroup$ – Willemien Jan 25 '14 at 13:30
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Integration by Parts: An Intuitive and Geometric Explanation by Sahand Rabbani.

EDIT: the cov formula.

The geometric idea is enclosed in the particular case $f(x)=k$ constant, $g(t)=pt+q$.

In this case, the cov formula $$(kp)(b-a)=\int_{a}^{b}kp\,dt = \int_{a}^{b}f(g(t))g'(t)\,dt = \int_{g(a)}^{g(b)}f(x)\,dx = \int_{g(a)}^{g(b)}k\,dx = k(g(b)-g(a))$$ says that two rectangles (what rectangles?) have the same area.

In the general case, approximation is required: supposing wlog $g$ increasing and taking a fine enough partition of the interval $[g(a),g(b)]$, in each subinterval $[t_k,t_{k+1}]$:

$f(g(t))\approx f(g(x_k))\qquad\qquad\qquad$ ($f\circ g$ is approx. constant because is continuous),

$g(t)\approx g'(t_k)(t-t_k)+g(t_k)\qquad\ \,$ ($g$ is approx. linear because is differentiable),

$g'(t)\approx g'(t_k)\qquad\qquad\qquad\ \ \ \ \ \ $ ($g'$ is approx. constant because is continuous).

And in each subinterval $[x_k,x_{k+1}]=[g(t_k),g(t_{k+1})]$:

$f(x)\approx f(x_k)\qquad\qquad\qquad\qquad$ ($f$ is approx. constant because is continuous).

Using the approximations:

$$ \int_{g(a)}^{g(b)}f(x)\,dx = \sum\int_{g(t_k)}^{g(t_{k+1})}f(x)\,dx \approx \sum\int_{g(t_k)}^{g(t_{k+1})}f(x_k)\,dx = $$

$$ \sum(g(t_{k+1})-g(t_k))f(x_k) \approx \sum(g'(t_k)(t_{k+1}-t_k)+g(t_k))-g(t_k))f(x_k) = $$

$$ \sum f(g(t_k))g'(t_k)(t_{k+1}-t_k)\approx \sum\int_{t_k}^{t_{k+1}}f(g(t))g'(t)\,dt = \int_{g(a)}^{g(b)}f(g(t))g'(t)\,dt. $$

This, done with $\epsilon-\delta$ rigor, will be a proof of the cov formula, but the geometric idea is the same that in the particular case.

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    $\begingroup$ I've seen the diagram before but I don't understand it. $u$ and $v$ do not have to be related in any way for integration by parts to work, but in the graph they're inverses... $\endgroup$ – dfg Jan 22 '14 at 23:27
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    $\begingroup$ @dfg This 2 page paper answers your a) question. I would suggest reading it slowly. If you understand that integration is the area under the curve, you might find the explanation odd because it is not rigorous, but it is how it works out. Maybe try it on graph paper on some simple function and count the squares to calculate both areas. $\endgroup$ – CAGT Jan 25 '14 at 14:20
  • $\begingroup$ This answers the "partial integration " part of the question, I would be most interested in the substitution part. Students (at least mine) generally understands partial integration well (its just the product formula of derivation "in reverse"), but have more difficulties with substitution. At least, in my last exam, many did the integration by parts Qs, but failed thye substitution ones, by trying to do them by partial integration and failing. Surprisimgly many failed in that way! $\endgroup$ – kjetil b halvorsen Jan 28 '14 at 11:33
  • $\begingroup$ @kjetil b halvorsen, see my answer. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 28 '14 at 11:43
  • $\begingroup$ @Martín-BlasPérezPinilla Are $p$ and $q$ constants? $\endgroup$ – dfg Jan 28 '14 at 17:47

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