Understanding Integration techniques?

Question

Could someone give me a geometric interpretation of:

a) Integration by Parts

b) Integration by Substitution

Thanks!

Answer 1

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.