The Fundamental Theorem of Calculus, Part 1: Unveiling the Derivative-Integral Link

1. The Integral as an "Accumulation" Function

So far, we have treated the definite integral ${\int }_a^{b}f(x)dx$ as a single number representing the area under a curve between two fixed points, $a$ and $b$.

But what if we let the upper limit be a variable? Let's define a new function, which we can call an area function or accumulation function, $A(x)$. This function measures the area under the curve of a function $f$ from a fixed starting point $a$ to a variable endpoint $x$.

We define it like this:

Notice we use $t$ for the variable inside the integral (the "dummy variable") to avoid confusion with the upper limit $x$.

The following video provides an excellent intuitive feel for this concept using the familiar analogy of velocity and distance.

As the video demonstrates, the area under the velocity graph from time 0 to time $T$ gives the total distance travelled. The function $s(T)={\int }_0^{T}v(t)dt$ is precisely one of these accumulation functions.

2. The Rate of Change of Area

Now for the crucial question: What is the derivative (or gradient) of this area function $A(x)$? In other words, what is $A^{\prime }(x)$?

Let's think about it intuitively. The derivative $A^{\prime }(x)$ represents the instantaneous rate at which the accumulated area is changing at the point $x$.

Consider increasing $x$ by a tiny amount, let's call it $dx$. The area function $A(x)$ will increase by a small amount, $dA$. This small increase in area, $dA$, is the area of a very thin sliver under the curve.

The sliver is almost a perfect rectangle.

• Its width is $dx$.
• Its height is approximately the value of the function at that point, $f(x)$.

So, the small change in area is approximately:

If we divide by $dx$, we get the rate of change of area with respect to $x$:

As the width $dx$ approaches zero, this approximation becomes exact. Therefore, the derivative of the area function is simply the original function itself!

This is the heart of FTC1. It tells us that differentiation "undoes" integration.

The next video provides a slightly more formal, but still highly visual, demonstration of this same conclusion.

3. The Fundamental Theorem of Calculus, Part 1

Let's formally state the theorem. It brings together all the ideas we've just explored.

As the text states, The Fundamental Theorem of Calculus, Part 1 says:
If $f$ is continuous on an interval $[a,b]$ and we define a function $F(x)={\int }_a^{x}f(t)dt$, then $F$ is differentiable and...
$F^{\prime }(x)=f(x)$
Or, written more directly:
$\frac{d}{dx}\left[{\int }_a^{x}f(t)dt\right]=f(x)$

This infographic provides a nice visual summary.

At its most basic level, applying FTC1 is straightforward. For example:
If $F(x)={\int }_2^x\cos (t)dt$, then $F^{\prime }(x)=\cos (x)$.
We simply replace the $t$ in the integrand with the $x$ from the upper limit.

4. Combining FTC1 with the Chain Rule

Things get more interesting when the upper limit of the integral is not just $x$, but a function of $x$, like $x^2$ or $\sin (x)$. In this case, we need to combine FTC1 with the Chain Rule you learned in a previous module.

Let's say we have $H(x)={\int }_a^{g(x)}f(t)dt$.
If we let $u=g(x)$, then $H(x)$ is a composition: $H(x)={\int }_a^{u}f(t)dt$.
By the Chain Rule, $H^{\prime }(x)=\frac{dH}{du}\cdot \frac{du}{dx}$.

• From FTC1, we know $\frac{dH}{du}=f(u)$.
• From the definition of $u$, we know $\frac{du}{dx}=g^{\prime }(x)$.

Substituting everything back, we get the general rule:
$\frac{d}{dx}\left[{\int }_a^{g(x)}f(t)dt\right]=f(g(x))\cdot g^{\prime }(x)$

In words: substitute the upper limit into the integrand, then multiply by the derivative of the upper limit.

Let's study the section of the textbook that walks through this.

5. Tying It All Together: How Part 1 Implies Part 2

We can now use FTC1 to understand why FTC2 works.

1. FTC1 tells us that the function $A(x)={\int }_a^{x}f(t)dt$ is an antiderivative of $f(x)$, because $A^{\prime }(x)=f(x)$.

2. We also know that any two antiderivatives of the same function can only differ by a constant. So if $F(x)$ is any other antiderivative of $f(x)$, it must be that $F(x)=A(x)+C$.

3. Now let's look at the expression $F(b)-F(a)$ from last lesson:

   $$F(b)-F(a)=(A(b)+C)-(A(a)+C)=A(b)-A(a)$$

4. What are $A(b)$ and $A(a)$? From our definition of $A(x)$:

   • $A(b)={\int }_a^{b}f(t)dt$
   • $A(a)={\int }_a^{a}f(t)dt=0$ (since the area from a point to itself is zero)

5. Substituting these back in, we get:

   $$F(b)-F(a)={\int }_a^{b}f(t)dt-0={\int }_a^{b}f(x)dx$$

This is exactly the formula for FTC2! This shows that the rule we used in the last lesson is a direct logical consequence of the fact that differentiation and integration are inverse operations.

The final part of the 3Blue1Brown video you watched earlier elegantly summarises this connection.

Conclusion

Today, you've explored the conceptual core of calculus. While the previous lesson gave you a practical tool (FTC2), this lesson provides the fundamental understanding that underpins it all.

Key Takeaways:

• An integral can be defined as an accumulation function $A(x)={\int }_a^{x}f(t)dt$, which measures the area under a curve up to a variable point $x$.
• The Fundamental Theorem of Calculus, Part 1 states that the derivative of this accumulation function is the original function: $A^{\prime }(x)=f(x)$.
• This establishes differentiation and integration as inverse processes. One "undoes" the other.
• This inverse relationship is the reason why we can use antiderivatives to calculate definite integrals, which is the statement of FTC2.
• When finding the derivative of an integral with a function as its upper limit, you must use the Chain Rule.

This lesson concludes our introduction to the core concepts of integration. In the next module, we will circle back to differentiation and begin exploring its powerful applications, such as finding the maximum and minimum values of functions and solving optimisation problems.