Understanding Integration techniques? Could someone give me a geometric interpretation of:
a) Integration by Parts
b) Integration by Substitution 
Thanks!
 A: 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.
