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Introduction to Calculus

Calculus is one of the most beautiful and powerful branches of mathematics. At its heart lies the Fundamental Theorem of Calculus, which establishes the relationship between differentiation and integration.

1. Derivatives

Definition 1.1(Derivative)
The derivative of a function ff at a point aa is defined as the limit:
f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}
provided this limit exists.

The derivative measures the instantaneous rate of change of a function. For a function y=f(x)y = f(x), we also write the derivative as dydx\frac{dy}{dx} or ddxf(x)\frac{d}{dx}f(x).

Example 1.1
Let f(x)=x2f(x) = x^2. Using the definition:
f′(x)=lim⁡h→0(x+h)2−x2h=lim⁡h→02xh+h2h=lim⁡h→0(2x+h)=2xf'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x + h) = 2x

2. Integrals

Definition 2.1(Definite Integral)
The definite integral of a function ff from aa to bb is:
∫abf(x) dx=lim⁡n→∞∑i=1nf(xi∗)Δx\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x
where Δx=b−an\Delta x = \frac{b-a}{n} and xi∗x_i^* is a sample point in the ii-th subinterval.

3. The Fundamental Theorem

Theorem 3.1(Fundamental Theorem of Calculus)
Let ff be continuous on [a,b][a, b]. Then:
  1. If F(x)=∫axf(t) dtF(x) = \int_a^x f(t) \, dt, then F′(x)=f(x)F'(x) = f(x).
  2. If FF is any antiderivative of ff, then:
    ∫abf(x) dx=F(b)−F(a)\int_a^b f(x) \, dx = F(b) - F(a)
Proof.
We prove the first part. Let F(x)=∫axf(t) dtF(x) = \int_a^x f(t) \, dt. Then:
F′(x)=lim⁡h→0F(x+h)−F(x)h=lim⁡h→01h∫xx+hf(t) dtF'(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} = \lim_{h \to 0} \frac{1}{h} \int_x^{x+h} f(t) \, dt
By the Mean Value Theorem for Integrals, there exists c∈[x,x+h]c \in [x, x+h] such that ∫xx+hf(t) dt=f(c)⋅h\int_x^{x+h} f(t) \, dt = f(c) \cdot h. As h→0h \to 0, we have c→xc \to x, and by continuity of ff, we get F′(x)=f(x)F'(x) = f(x).
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Remark. The Fundamental Theorem of Calculus shows that differentiation and integration are inverse operations, providing a powerful tool for evaluating definite integrals.
Example 3.1
Evaluate ∫02x2 dx\int_0^2 x^2 \, dx.

Solution: Since ddx(x33)=x2\frac{d}{dx}\left(\frac{x^3}{3}\right) = x^2, we have:

∫02x2 dx=[x33]02=83−0=83\int_0^2 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3}