Calculus I: Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is not just another formula to memorize; it is the conceptual linchpin that connects the two major branches of calculus—differentiation and integration. For engineering, this connection is essential, transforming integration from a painstaking limit-sum process into a powerful tool for analyzing accumulated change, from computing the total force on a surface to determining the net displacement of a system. Without the FTC, the practical application of calculus to solve real-world problems would be nearly impossible.

The Bridge Between Two Worlds

To appreciate the FTC, you must first understand the two seemingly separate problems it unites. Differentiation is the mathematics of instantaneous change. Given a position function, its derivative tells you the velocity. Integration, specifically the definite integral, is the mathematics of accumulation. Given a velocity function, the definite integral over a time interval tells you the net displacement. Before the FTC, these were separate operations requiring different techniques. The FTC reveals they are inverse processes. This means if you accumulate a rate of change, you get back the net change in the original quantity. This inverse relationship is the engine behind virtually every calculus application in physics and engineering.

FTC Part 1: The Derivative of an Integral

The first part of the theorem, often called the "derivative form," establishes that differentiation "undoes" integration. It states:

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

In simpler terms, if you create a new function $F(x)$ by integrating a continuous function $f(t)$ from a fixed starting point $a$ to a variable endpoint $x$, then the derivative of this area-accumulation function is simply the original function evaluated at $x$. The upper limit $x$ is the crucial variable.

Example: Let $f(t)=2t+1$. Define $F(x)={\int }_0^x(2t+1)dt$. FTC Part 1 tells us immediately that $F^{\prime }(x)=2x+1$, without needing to first evaluate the integral. To check, we can compute the integral: $F(x)=[t^2+t{]}_0^x=x^2+x$. Differentiating this gives $2x+1$, confirming the theorem.

For engineering, this is profound. If $f(t)$ represents a rate (e.g., the rate of fluid flow into a tank), then $F(x)$ represents the total amount accumulated up to time $x$. FTC Part 1 confirms that the rate of change of the total amount is exactly the flow rate at the present moment.

FTC Part 2: The Evaluation Theorem

The second part, the "evaluation form," is the workhorse for computing definite integrals. It states:

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ (that is, $F^{\prime }=f$), then $${\int }_a^{b}f(x)dx=F(b)-F(a).$$

This theorem solves the central practical problem of integration. Instead of calculating the limit of a Riemann sum, you find a function $F$ whose derivative is $f$, and then simply evaluate $F$ at the boundaries. The notation $F(x){|}_a^b$ or $[F(x){]}_a^b$ is shorthand for $F(b)-F(a)$.

Step-by-Step Application:

1. Identify: Ensure the integrand $f(x)$ is continuous on the interval $[a,b]$.
2. Find an Antiderivative: Determine a function $F(x)$ such that $F^{\prime }(x)=f(x)$. The constant of integration $C$ is not needed as it cancels out.
3. Evaluate: Compute the difference $F(b)-F(a)$.

Example: Compute ${\int }_1^4(3x^2+\sqrt{x})dx$.

1. The function is continuous on $[1,4]$.
2. Find an antiderivative. Rewrite $\sqrt{x}$ as $x^{1/2}$.

$$F(x)=x^3+\frac{x^{3/2}}{3/2}=x^3+\frac{2}{3}{x}^{3/2}.$$

1. Evaluate:

$$[F(x){]}_1^4=\left(4^3+\frac{2}{3}(4{)}^{3/2}\right)-\left(1^3+\frac{2}{3}(1{)}^{3/2}\right)=\left(64+\frac{2}{3}(8)\right)-\left(1+\frac{2}{3}\right)=\left(64+\frac{16}{3}\right)-\left(\frac{5}{3}\right).$$ This simplifies to $64+\frac{11}{3}=64+3.\overline{6}=67.\overline{6}$.

A Sketch of the Proofs

Understanding the logic behind the FTC solidifies your grasp of the concepts.

FTC Part 1 Proof Sketch: We want to show $F^{\prime }(x)=f(x)$, where $F(x)={\int }_a^{x}f(t)dt$. By the definition of the derivative: $$F^{\prime }(x)=\underset{h\to 0}{\lim }\frac{F(x+h)-F(x)}{h}.$$ The numerator $F(x+h)-F(x)$ is the difference in area from $a$ to $x+h$ and from $a$ to $x$, which is precisely the area from $x$ to $x+h$. That is: $$F(x+h)-F(x)={\int }_x^{x+h}f(t)dt.$$ For a very small $h$, this thin slice of area is approximately a rectangle with height $f(x)$ and width $h$. Therefore, $\frac{F(x+h)-F(x)}{h}\approx \frac{f(x)\cdot h}{h}=f(x)$. As $h\to 0$, this approximation becomes exact, proving $F^{\prime }(x)=f(x)$.

FTC Part 2 Proof Sketch: Let $F(x)={\int }_a^{x}f(t)dt$ be the specific antiderivative from Part 1. Let $G$ be any other antiderivative of $f$ (so $G^{\prime }=f$). Since $F$ and $G$ have the same derivative, they differ by at most a constant: $G(x)=F(x)+C$. Now, evaluate the definite integral: $${\int }_a^{b}f(t)dt=F(b)=F(b)-0.$$ But $F(a)={\int }_a^{a}f(t)dt=0$. So, ${\int }_a^{b}f(t)dt=F(b)-F(a)$. Since $G(x)=F(x)+C$, we have $G(b)-G(a)=[F(b)+C]-[F(a)+C]=F(b)-F(a)$. Thus, any antiderivative $G$ works for the evaluation.

Engineering Applications and Why FTC is Supreme

The FTC's importance cannot be overstated. It is the "fundamental" theorem because it makes calculus a coherent, applicable system.

• From Rate to Total: Given a known rate function (e.g., velocity $v(t)$, current $i(t)$, power $P(t)$), you can find the net change (displacement, charge, energy) over an interval instantly: $\text{Net Change}={\int }_{t_1}^{t_2}\text{Rate}(t)dt$.
• Simplifying Computations: It reduces the geometric problem of finding areas under curves (or volumes, work, etc.) to the algebraic problem of finding antiderivatives.
• Modeling Systems: In dynamics, the FTC underpins the relationship between force and momentum (impulse-momentum theorem) and between power and work.

Without the FTC, you would be stuck calculating sums of infinitesimals for every problem. With it, you have a direct, computational pathway that is essential for modeling everything from signal processing to thermodynamics.

Common Pitfalls

1. Misapplying FTC Part 1 with a Composite Upper Limit: A classic trap is a function like $H(x)={\int }_0^{x^2}\sin (t)dt$. Here, the upper limit is $x^2$, not $x$. You must apply the Chain Rule. If we let $u=x^2$, then $H^{\prime }(x)=\frac{d}{du}\left({\int }_0^u\sin tdt\right)\cdot \frac{du}{dx}=\sin (u)\cdot 2x=2x\sin (x^2)$.

1. Using the Wrong Antiderivative: The antiderivative $F(x)$ in FTC Part 2 must be differentiable (and hence continuous) on the entire closed interval $[a,b]$. For an integral like ${\int }_{-1}^1\frac{1}{x^2}dx$, the integrand is undefined at $x=0$. FTC Part 2 does not apply directly; this is an improper integral requiring a limit process.

1. Forgetting the Constant of Integration When It's Needed (and Vice Versa): When using FTC Part 2 to evaluate a definite integral, you do not need the "+ C" because it cancels. However, if you are using integration to solve for a general function (e.g., solving a differential equation), the constant of integration is essential.

1. Confusing the Variable of Integration: In $F(x)={\int }_a^{x}f(t)dt$, the variable $t$ is a "dummy variable." It does not appear in the final answer for $F(x)$. You could use any symbol except $x$. When differentiating $F(x)$, the result is $f(x)$, not $f(t)$.

Summary

• The Fundamental Theorem of Calculus unifies differentiation and integration, showing they are inverse operations.
• FTC Part 1 states that the derivative of an integral with a variable upper limit is the integrand evaluated at that limit: $\frac{d}{dx}{\int }_a^{x}f(t)dt=f(x)$. Remember the Chain Rule if the upper limit is a function of $x$.
• FTC Part 2 (Evaluation Theorem) provides the primary method for computing definite integrals: ${\int }_a^{b}f(x)dx=F(b)-F(a)$, where $F$ is any antiderivative of $f$.
• This theorem transforms integration from a limiting sum into a tractable algebraic evaluation, making it indispensable for calculating net change from a rate in all engineering fields.
• Always verify the function's continuity on the interval and carefully handle the limits of integration, especially when they are functions themselves.