Let us first recall the proposition mentioned (without proof) in yesterday's post:

$$

F'(x) = \big( \int_{a}^{x}f(t)dt \big)' = f(x)

$$

for all $x \in \Delta$.

If $x \in \Delta$ is a variable point,

then the value of the definite integral $F(x)= \int_{a}^{x}f(t)dt$ represents the area of the shaded region shown in the figure:

Considering $h$ to be infinitesimal, we can now compute

$$

\begin{array}{c}

\Delta F(x) = F(x+h)-F(x) = \int_{a}^{x+h}f(t)dt - \int_{a}^{x}f(t)dt = \\

\\

= \int_{x}^{x+h}f(t)dt = E(\Omega) \approx h \cdot f(x)

\end{array}

$$

In the above, the change $\Delta F(x)$ in the value of the function $F(x)=\int_{a}^{x}f(t)dt$ when $x$ changes to $x+h$ is denoted by $E(\Omega)$ and equals the area of the shaded strip $\Omega$, displayed in the next figure:

Consequently, $f(x) \approx \frac{E(\Omega)}{h}$ and since $h$ is considered to be infinitesimal, we can write:

$$

f(x)=\lim_{h \rightarrow 0}\frac{F(x+h)-F(x)}{h}=F'(x) = \frac{d \big(\int_{a}^{x}f(t)dt \big)}{dx}

$$

$$

G(x) = \int_{a}^{g(x)}f(t)dt

$$

The domain $D_{G}$ of $G$ will be:

$$

D_{G} = \{ x \in D_{g} \ , \ g(x) \in \Delta \}

$$

The composite function $G=F \circ g$ is differentiable on $\Delta_{2} = \Delta_{1} \cap D_{G}$. Its derivative can be computed by combining the above differentiation rule with the

$$

\Big( \int_{a}^{g(x)}f(t)dt \Big)' = f\big( g(x) \big) \cdot g'(x)

$$

for all $x \in \Delta_{2}$.

**Proposition:**Let a real function $f$, with domain $D_{f}$ and continuous on an interval $\Delta \subseteq D_{f} \subseteq \mathbb{R}$ and let $a \in \Delta$ be a fixed point of $\Delta$. Then the function $F(x)=\int_{a}^{x}f(t)dt$ is an antiderivative function of $f$. In other words:$$

F'(x) = \big( \int_{a}^{x}f(t)dt \big)' = f(x)

$$

for all $x \in \Delta$.

If $x \in \Delta$ is a variable point,

then the value of the definite integral $F(x)= \int_{a}^{x}f(t)dt$ represents the area of the shaded region shown in the figure:

**Remark:**Notice that the function $F(x)$ is defined on any interval $\Delta \subseteq D_{f}$ in which:- $a \in \Delta$
- f is continuous on $\Delta$

Considering $h$ to be infinitesimal, we can now compute

$$

\begin{array}{c}

\Delta F(x) = F(x+h)-F(x) = \int_{a}^{x+h}f(t)dt - \int_{a}^{x}f(t)dt = \\

\\

= \int_{x}^{x+h}f(t)dt = E(\Omega) \approx h \cdot f(x)

\end{array}

$$

In the above, the change $\Delta F(x)$ in the value of the function $F(x)=\int_{a}^{x}f(t)dt$ when $x$ changes to $x+h$ is denoted by $E(\Omega)$ and equals the area of the shaded strip $\Omega$, displayed in the next figure:

Consequently, $f(x) \approx \frac{E(\Omega)}{h}$ and since $h$ is considered to be infinitesimal, we can write:

$$

f(x)=\lim_{h \rightarrow 0}\frac{F(x+h)-F(x)}{h}=F'(x) = \frac{d \big(\int_{a}^{x}f(t)dt \big)}{dx}

$$

**Remarks:****(1).**The above should not be considered to be a rigorous proof. It should rather be taken as an intuitive line of thinking, aiming to shed some light into "what is really going on" inside the function $\int_{a}^{x}f(t)dt$**(2).**Let a real function $f$, continuous on an interval $\Delta \subseteq \mathbb{R}$ and let $a \in \Delta$ be a fixed point of $\Delta$. Let another real function $g(x)$ with domain $D_{g}$ and differentiable (and thus: continuous) on an interval $\Delta_{1} \subseteq D_{g}$. We can then consider the composite function $G=F \circ g$:$$

G(x) = \int_{a}^{g(x)}f(t)dt

$$

The domain $D_{G}$ of $G$ will be:

$$

D_{G} = \{ x \in D_{g} \ , \ g(x) \in \Delta \}

$$

The composite function $G=F \circ g$ is differentiable on $\Delta_{2} = \Delta_{1} \cap D_{G}$. Its derivative can be computed by combining the above differentiation rule with the

*chain rule of differentiation*(for differentiating composite functions). We readily get the following formula:$$

\Big( \int_{a}^{g(x)}f(t)dt \Big)' = f\big( g(x) \big) \cdot g'(x)

$$

for all $x \in \Delta_{2}$.