AP Calculus AB: Fundamental Theorem of Calculus Part 1

The Fundamental Theorem of Calculus is the bridge that unites the two major concepts of calculus—differentiation and integration. Understanding Fundamental Theorem of Calculus Part 1 (FTC 1) is crucial because it transforms how you solve problems, turning complex limit definitions of derivatives into straightforward evaluations. For engineers and scientists, this theorem is the mathematical backbone for modeling systems where a rate of change is connected to an accumulated quantity, such as relating velocity to total distance traveled.

Understanding the Accumulation Function

To grasp FTC 1, you must first understand the accumulation function. This is a function defined by a definite integral where one of the limits of integration is a variable, typically $x$. The classic form is: $$F(x)={\int }_a^{x}f(t)dt.$$ Here, $a$ is a constant (the lower limit of integration), $x$ is the independent variable (the upper limit of integration), and $t$ is a dummy variable of integration. The function $F(x)$ represents the net signed area under the curve of $f(t)$ from $t=a$ to $t=x$. As $x$ increases or decreases, the amount of accumulated area changes. Think of it like a car's odometer: if $f(t)$ represents speed, then $F(x)$ tells you the total distance traveled since time $t=a$.

The Formal Statement of FTC Part 1

The Fundamental Theorem of Calculus, Part 1 provides a stunningly simple way to find the derivative of an accumulation function. It states:

If $f$ is continuous on an interval $[a,b]$, and if $F$ is defined by $$F(x)={\int }_a^{x}f(t)dt$$ for all $x$ in $[a,b]$, then $F$ is differentiable on $(a,b)$ and $$F^{\prime }(x)=f(x).$$

In plain language: The derivative of an integral with a variable upper limit is just the original function evaluated at that upper limit. This formally establishes that differentiation and integration are inverse processes. Differentiation "undoes" integration, and vice versa (as explored in FTC Part 2).

Applying the Theorem: Finding Derivatives of Integrals

The direct application of FTC 1 is to find derivatives of functions defined by integrals. The process is a direct substitution, provided the variable is the upper limit.

Example 1: Find $F^{\prime }(x)$ if $F(x)={\int }_2^x(t^3+5)dt$. Since $x$ is the upper limit and the integrand $(t^3+5)$ is continuous, FTC 1 applies directly. You simply replace the dummy variable $t$ with the upper limit $x$: $$F^{\prime }(x)=x^3+5.$$

The power of FTC 1 extends to more complex upper limits. If the upper limit is a function of $x$, say $u(x)$, you must apply the Chain Rule.

Example 2: Find the derivative of $G(x)={\int }_1^{\sin x}\sqrt{t^4+1}dt$. Here, the upper limit is $u(x)=\sin x$. FTC 1 combined with the Chain Rule gives: $$G^{\prime }(x)=f(u(x))\cdot u^{\prime }(x)=\sqrt{(\sin x{)}^4+1}\cdot \cos x.$$ The process is: 1) Evaluate the integrand at the upper limit function, 2) Multiply by the derivative of that upper limit function.

Handling Variable Lower Limits and Swapped Bounds

What if the variable is in the lower limit? You can manipulate the integral's properties to use FTC 1. Recall that swapping the bounds of integration introduces a negative sign: $${\int }_a^{x}f(t)dt=-{\int }_x^{a}f(t)dt.$$

Example 3: Find $H^{\prime }(x)$ if $H(x)={\int }_x^4{e}^{t^2}dt$. First, swap the bounds to make $x$ the upper limit: $$H(x)=-{\int }_4^x{e}^{t^2}dt.$$ Now apply FTC 1: $$H^{\prime }(x)=-e^{x^2}.$$

The most general form accounts for both limits being functions. For $F(x)={\int }_{v(x)}^{u(x)}f(t)dt$, where $v(x)$ and $u(x)$ are differentiable and $f$ is continuous, the derivative is: $$F^{\prime }(x)=f(u(x))\cdot u^{\prime }(x)-f(v(x))\cdot v^{\prime }(x).$$ This incorporates the effect of a variable lower limit by subtracting its contribution, following from the property of splitting integrals and applying the Chain Rule.

Common Pitfalls

1. Forgetting the Chain Rule with Composite Upper/Lower Limits. The most frequent error is applying FTC 1 directly to an integral like ${\int }_a^{g(x)}f(t)dt$ and writing $f(x)$ instead of $f(g(x))\cdot g^{\prime }(x)$. Always check: is the limit simply $x$, or is it a function of $x$?

1. Misapplying the Sign with Variable Lower Limits. If the variable is in the lower limit and you don't correctly swap bounds or use the general formula, you will get the sign wrong. Remember, if $F(x)={\int }_x^{a}f(t)dt$, then $F^{\prime }(x)=-f(x)$.

1. Differentiating the Integrand Instead of Applying FTC. A fundamental misunderstanding is trying to differentiate $f(t)$ with respect to $x$. The variable of integration is $t$; the integral's value depends on $x$ only through the limits. FTC 1 gives you the answer without needing to find the antiderivative first.

1. Overlooking the Continuity Requirement. FTC 1 requires $f$ to be continuous on the interval containing $x$. If $f$ has a discontinuity at the point where you are evaluating the derivative, the theorem does not apply, and the accumulation function may not be differentiable there.

Summary

• FTC Part 1 states that for a continuous function $f$, the derivative of the accumulation function $F(x)={\int }_a^{x}f(t)dt$ is $F^{\prime }(x)=f(x)$. This crystallizes the inverse relationship between differentiation and integration.
• The primary application is finding derivatives of functions defined by integrals. If the upper limit is $x$, simply evaluate the integrand at $x$.
• When the limit(s) of integration are functions of $x$, you must combine FTC 1 with the Chain Rule. The general formula is $\frac{d}{dx}{\int }_{v(x)}^{u(x)}f(t)dt=f(u(x))u^{\prime }(x)-f(v(x))v^{\prime }(x)$.
• Always verify the continuity of the integrand and correctly handle the sign when the variable is in the lower limit of integration to avoid common calculation errors.