For a differentiable function $f:\mathbb{R}\rightarrow\mathbb{R}$, using integration by parts, we get
$$\frac{(b-a)(f(a)+f(b))}{2}-\int_a^bf(x)dx=\int_a^b\left(x-\frac{a+b}{2}\right)f'(x)dx$$
The LHS represents the error of the trapezoidal approximation of the integral. Is there an geometric interpretation of the expression
$$\int_a^b\left(x-\frac{a+b}{2}\right)f'(x)dx$$
Best Answer
Let $y =f(x)$ to rewrite the expression as
\begin{align} \int_a^b\left(x-\frac{a+b}{2}\right)f'(x)dx =&\int_{f(b)}^{f(b)} \left(f^{-1}(y)-\frac{a+b}{2}\right)dy\\= &\int_{f(a)}^{f(b)} f^{-1}(y)dy - \frac{(a+b)(f(b)-f(a))}{2} \end{align} Thus, geometrically, the error of the trapezoidal approximation for the $x$-integration is equal to the error of the rectangular approximation for the $y$-integration.