Integrating strips normal to the curve How do you go about finding the following area under the curve where strips are supposed to be normal to the curve?

 A: With a normal integration and adding and removing the triangles at each extreme we have
$$A=\int_a^bf(x)dx+\frac12f(b)^2f'(b)-\frac12f(a)^2f'(a)$$
Let's keep this formula for testing.
If we take a point $x\in(a,b)$ and move $dx$, then the area of that segment will be:
\begin{align}
dS &= \frac12f(x+dx)^2f'(x+dx) + f(x)dx - \frac12f(x)^2f'(x)
\\&=\frac{f(x+dx)^2f'(x+dx)-f(x)^2f'(x)}2+f(x)dx 
\\&=\frac{f(x+dx)^2f'(x+dx)-f(x)^2f'(x)}{dx}\cdot\frac{dx}2+f(x)dx
\\&=\frac{d}{dx}\big(f(x)^2f'(x)\big)\frac{dx}2+f(x)dx & \text{(1)}
\\&=\big(2f(x)f'(x)^2+f(x)^2f''(x)\big)\frac{dx}2+f(x)dx
\\&=f(x)f'(x)^2dx+\frac12f(x)^2f''(x)dx+f(x)dx
\end{align}
From (1)
\begin{align}
A=\int_SdS&=\int_a^b\left(\frac12\frac{d}{dx}\big(f(x)^2f'(x)\big)dx+f(x)dx\right)
\\&=\frac12\int_a^b\frac{d}{dx}\big(f(x)^2f'(x)\big)dx+\int_a^bf(x)dx
\\&=\frac12\left.f(x)^2f'(x)\right|_a^b+\int_a^bf(x)dx
\\&=\frac{f(b)^2f'(b)-f(a)^2f'(a)}2+\int_a^bf(x)dx
\end{align}
Which is the same formula confirming the result:
\begin{align}
dS &= f(x)f'(x)^2dx+\frac12f(x)^2f''(x)dx+f(x)dx
\\&= f(x)\big(f'(x)^2+\frac12f(x)f''(x)+1\big)dx
\end{align}
