If the function f(x) satisfies the conditions

If the function $f(x)$ satisfies the conditions of the Newton-Leibniz theorem (Fundamental Theorem of Calculus) for $x \in [a, b]$, prove that: $$ \frac{d}{dx}\left(\int_a^x f(t) \, dt\right) = f(x) $$

This is one of the most important theorems in calculus, establishing the relationship between differentiation and integration.

Statement of the Fundamental Theorem of Calculus (Part 1)

Fundamental Theorem of Calculus - Part 1

Let $f$ be a continuous function on $[a, b]$. Define the function $F$ by:

$$F(x) = \int_a^x f(t) \, dt \quad \text{for } x \in [a, b]$$

Then $F$ is differentiable on $(a, b)$ and:

$$F'(x) = f(x) \quad \text{for all } x \in (a, b)$$

Proof

Let $F(x) = \int_a^x f(t) \, dt$. We need to prove that $F'(x) = f(x)$.

Step 1: Definition of the derivative

By definition, the derivative of $F$ at $x$ is:

$$F'(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h}$$

provided this limit exists.

Step 2: Express $F(x+h)$ and $F(x)$

Using the definition of $F$:

$$F(x+h) = \int_a^{x+h} f(t) \, dt$$

$$F(x) = \int_a^x f(t) \, dt$$

Step 3: Compute the difference $F(x+h) - F(x)$

Using the additive property of integrals:

$$F(x+h) - F(x) = \int_a^{x+h} f(t) \, dt - \int_a^x f(t) \, dt = \int_x^{x+h} f(t) \, dt$$

This follows from the property:

$$\int_a^{x+h} f(t) \, dt = \int_a^x f(t) \, dt + \int_x^{x+h} f(t) \, dt$$

Step 4: Form the difference quotient

$$\frac{F(x+h) - F(x)}{h} = \frac{1}{h} \int_x^{x+h} f(t) \, dt$$

Step 5: Apply the Mean Value Theorem for Integrals

Since $f$ is continuous on $[x, x+h]$ (or $[x+h, x]$ if $h < 0$), by the Mean Value Theorem for Integrals, there exists a point $c$ between $x$ and $x+h$ such that:

$$\int_x^{x+h} f(t) \, dt = f(c) \cdot h$$

where $c \in [x, x+h]$ if $h > 0$, or $c \in [x+h, x]$ if $h < 0$.

Step 6: Substitute into the difference quotient

$$\frac{F(x+h) - F(x)}{h} = \frac{f(c) \cdot h}{h} = f(c)$$

where $c$ is between $x$ and $x+h$.

Step 7: Take the limit as $h \to 0$

As $h \to 0$:

• The point $c$ is squeezed between $x$ and $x+h$
• Therefore, $c \to x$ as $h \to 0$
• Since $f$ is continuous at $x$, we have $f(c) \to f(x)$ as $c \to x$

Thus:

$$F'(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} = \lim_{h \to 0} f(c) = \lim_{c \to x} f(c) = f(x)$$

The last equality holds because $f$ is continuous at $x$.

Conclusion

We have proven that:

$$\boxed{\frac{d}{dx}\left(\int_a^x f(t) \, dt\right) = f(x)}$$

Alternative proof using the definition directly

Here's a more direct approach for the case when $h > 0$ is small:

Since $f$ is continuous at $x$, for any $\varepsilon > 0$, there exists $\delta > 0$ such that:

$$|t - x| < \delta \implies |f(t) - f(x)| < \varepsilon$$

For $0 < h < \delta$, all $t \in [x, x+h]$ satisfy $|t - x| < \delta$, so:

$$f(x) - \varepsilon < f(t) < f(x) + \varepsilon$$

Integrating over $[x, x+h]$:

$$\int_x^{x+h} (f(x) - \varepsilon) \, dt < \int_x^{x+h} f(t) \, dt < \int_x^{x+h} (f(x) + \varepsilon) \, dt$$

$$h(f(x) - \varepsilon) < \int_x^{x+h} f(t) \, dt < h(f(x) + \varepsilon)$$

Dividing by $h > 0$:

$$f(x) - \varepsilon < \frac{1}{h}\int_x^{x+h} f(t) \, dt < f(x) + \varepsilon$$

Therefore:

$$\left|\frac{F(x+h) - F(x)}{h} - f(x)\right| = \left|\frac{1}{h}\int_x^{x+h} f(t) \, dt - f(x)\right| < \varepsilon$$

Since $\varepsilon$ was arbitrary, we conclude:

$$\lim_{h \to 0^+} \frac{F(x+h) - F(x)}{h} = f(x)$$

A similar argument works for $h \to 0^-$, proving $F'(x) = f(x)$.

Geometric interpretation

The function $F(x) = \int_a^x f(t) \, dt$ represents the accumulated area under the curve $y = f(t)$ from $t = a$ to $t = x$.

The derivative $F'(x)$ represents the rate of change of this accumulated area.

When we increase $x$ by a small amount $h$, the additional area is approximately $f(x) \cdot h$ (a thin rectangle of height $f(x)$ and width $h$).

Therefore, the rate of change of area is:

$$F'(x) = \lim_{h \to 0} \frac{\text{additional area}}{h} = \lim_{h \to 0} \frac{f(x) \cdot h}{h} = f(x)$$

Important remarks

1. Conditions required: The key condition is that $f$ must be continuous on $[a, b]$. This ensures that the integral exists and that we can apply the Mean Value Theorem for Integrals.
2. Variable limits: This theorem shows that integration (from a constant $a$ to a variable $x$) and differentiation (with respect to $x$) are inverse operations in a certain sense.
3. Lower limit doesn't matter for the derivative: Notice that the choice of $a$ doesn't affect the derivative. Whether we use $\int_0^x f(t) \, dt$ or $\int_5^x f(t) \, dt$, the derivative is still $f(x)$.
4. Connection to Part 2 of FTC: Combined with the second part of the Fundamental Theorem (which states that if $F'(x) = f(x)$, then $\int_a^b f(x) \, dx = F(b) - F(a)$), this provides the complete relationship between derivatives and integrals.

Example application

Let $f(t) = t^2$ and consider $F(x) = \int_0^x t^2 \, dt$.

By direct calculation:

$$F(x) = \int_0^x t^2 \, dt = \left[\frac{t^3}{3}\right]_0^x = \frac{x^3}{3}$$

By the Fundamental Theorem:

$$F'(x) = f(x) = x^2$$

We can verify: $F'(x) = \frac{d}{dx}\left(\frac{x^3}{3}\right) = x^2$ ✓

Key lesson: The Fundamental Theorem of Calculus establishes that differentiation and integration are inverse operations, and provides a powerful tool for evaluating integrals and understanding the relationship between rates of change and accumulation.