Let \(F(x)\) be a differentiable function. We’d like to show that

\[\int_a^b \bigg( \frac{d}{dx} F(x) \bigg)\,dx = F(b) - F(a)\]

Expanding the definition of the derivative and assuming we can swap limits and integrals, we would have

\[\begin{align*} &\int_a^b \bigg( \frac{d}{dx} F(x) \bigg)\,dx \\ &= \int_a^b \lim_{h \to 0} \frac{F(x+h) - F(x)}{h}\,dx\\ &= \lim_{h \to 0} \frac{1}{h} \bigg(\int_a^b F(x+h) \,dx - \int_a^b F(x) \,dx \bigg)\\ \end{align*}\]

For example with \(F(x) = \sin(x)\), \(a=0\), \(b=\pi/6\) and \(h=-0.1\), we would have \(\int_a^b F(x+h) \,dx - \int_a^b F(x) \,dx\) be the red shaded region below. The red curve is \(\sin x\) while the blue curve is \(\sin(x+h)\).

Notice that the red shaded region is very similar to the blue shaded region.

As \(h\) and \(b-a\) tends to 0, the blue shaded region looks more and more like a parellogram with height \(F(b) - F(a)\) and width \(h\) which has area \(h \times (F(b)- F(a))\) as such

\[\begin{align*} &\int_a^b \bigg( \frac{d}{dx} F(x) \bigg)\,dx \\ &= \lim_{h \to 0} \frac{1}{h} \bigg(\int_a^b F(x+h) \,dx - \int_a^b F(x) \,dx \bigg)\\ &= \lim_{h \to 0} \frac{1}{h} h \times (F(b) - F(a)) \\ &= F(b) - F(a) \end{align*}\]


The advantage of this visualization is that we dealt with the differences of the area integrals more directly. The disadvantage is that we omitted using the intuition of the derivative / slope of \(F(x)\) all together.