My Calculus Yap

by jasonnel | posted January 14th, 2025


Exploring the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is one of the most profound results in mathematics, bridging the concepts of differentiation and integration. This post will explore the theorem in detail, providing intuition, formal statements, and examples.


The Two Parts of the Fundamental Theorem of Calculus

Part 1: Differentiation of the Integral

The first part of the FTC states that if ff is a continuous function on an interval [a,b][a, b], and we define a new function F(x)F(x) as:

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

then Fβ€²(x)=f(x)F'(x) = f(x) for all x∈[a,b]x \in [a, b]. In other words, the derivative of the integral of ff gives back ff itself.

This result provides a deep connection between differentiation and integration. Let’s break this down:

  1. Intuition: The integral accumulates the "area under the curve" of f(t)f(t) from aa to xx. Differentiating this accumulation with respect to xx gives back the instantaneous rate of change of ff at xx, which is simply f(x)f(x).

  2. Key Assumption: The function ff must be continuous on the interval [a,b][a, b] for this result to hold.


Part 2: The Evaluation Theorem

The second part of the FTC states that if ff is continuous on [a,b][a, b] and FF is any antiderivative of ff (i.e., Fβ€²(x)=f(x)F'(x) = f(x)), then:

∫abf(x) dx=F(b)βˆ’F(a).\int_a^b f(x) \, dx = F(b) - F(a).

This result simplifies the computation of definite integrals by allowing us to evaluate the antiderivative at the endpoints of the interval.

  1. Intuition: Instead of summing up infinitesimal areas under the curve directly, we can simply find an antiderivative FF and compute the difference F(b)βˆ’F(a)F(b) - F(a).

  2. Key Assumption: Again, the continuity of ff on [a,b][a, b] is crucial.


A Proof Sketch for the First Part

To justify the first part of the FTC, let’s consider:

F(x)=∫axf(t) dt.F(x) = \int_a^x f(t) \, dt.

The derivative of F(x)F(x) is given by the definition of the derivative:

Fβ€²(x)=lim⁑hβ†’0F(x+h)βˆ’F(x)h.F'(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h}.

Expanding F(x+h)F(x+h) and F(x)F(x) using the definition of FF, we have:

F(x+h)βˆ’F(x)=∫ax+hf(t) dtβˆ’βˆ«axf(t) dt.F(x+h) - F(x) = \int_a^{x+h} f(t) \, dt - \int_a^x f(t) \, dt.

Using the additivity property of definite integrals:

∫ax+hf(t) dt=∫axf(t) dt+∫xx+hf(t) dt,\int_a^{x+h} f(t) \, dt = \int_a^x f(t) \, dt + \int_x^{x+h} f(t) \, dt,

we find that:

F(x+h)βˆ’F(x)=∫xx+hf(t) dt.F(x+h) - F(x) = \int_x^{x+h} f(t) \, dt.

Dividing by hh:

F(x+h)βˆ’F(x)h=1h∫xx+hf(t) dt.\frac{F(x+h) - F(x)}{h} = \frac{1}{h} \int_x^{x+h} f(t) \, dt.

As h→0h \to 0, the interval [x,x+h][x, x+h] shrinks, and since ff is continuous, the average value of ff on this interval converges to f(x)f(x). Thus:

lim⁑hβ†’01h∫xx+hf(t) dt=f(x),\lim_{h \to 0} \frac{1}{h} \int_x^{x+h} f(t) \, dt = f(x),

which implies:

Fβ€²(x)=f(x).F'(x) = f(x).


Examples

Example 1: A Simple Integral

Let f(x)=2xf(x) = 2x. Compute ∫032x dx\int_0^3 2x \, dx using the FTC.

  1. Find an antiderivative: F(x)=x2F(x) = x^2.
  2. Apply the evaluation theorem:

∫032x dx=F(3)βˆ’F(0)=32βˆ’02=9.\int_0^3 2x \, dx = F(3) - F(0) = 3^2 - 0^2 = 9.


Example 2: Differentiating an Integral

Let f(t)=sin⁑(t)f(t) = \sin(t) and define F(x)=∫0xsin⁑(t) dtF(x) = \int_0^x \sin(t) \, dt. Compute Fβ€²(x)F'(x).

By the first part of the FTC:

Fβ€²(x)=sin⁑(x).F'(x) = \sin(x).


Applications and Insights

The FTC is essential in various fields:

  1. Physics: Calculating work done, displacement, and energy often involves definite integrals.
  2. Engineering: Analysis of systems, such as finding the response of circuits or mechanical systems, uses integrals.
  3. Probability: The cumulative distribution function (CDF) of a random variable is an integral, and its derivative gives the probability density function (PDF).

Conclusion

The Fundamental Theorem of Calculus is a cornerstone of mathematical analysis, providing a bridge between two seemingly distinct concepts: differentiation and integration. By understanding both parts of the theorem, one gains powerful tools to solve a wide range of problems in mathematics, science, and engineering.

If you’ve made it this far, try applying the FTC to your own examples! For instance:

  • Compute ∫14(3x2βˆ’2x+1) dx\int_1^4 (3x^2 - 2x + 1) \, dx.
  • Differentiate G(x)=∫2xeβˆ’t2 dtG(x) = \int_2^x e^{-t^2} \, dt.

Feel free to share your solutions and thoughts below!