MA 16500 Notes | Hanh Vo

Goal: Understand the Fundamental Theorem of Calculus (both parts) and its significance in connecting derivatives and integrals.

1. Introduction

1.1 Statement

Let \(f\) be continuous on \([a, b]\).

Part 1 (Derivative of the Integral):

Define

\[ A(x) = \int_a^x f(t)\, dt \]

Then \(A\) is an antiderivative of \(f\), i.e.,

\[ A'(x) = f(x) \]

or equivalently,

\[ \frac{d}{dx} \Big( \int_a^x f(t)\, dt \Big) = f(x) \]

Part 2 (Evaluation using any antiderivative):

Let \(F(x)\) be any antiderivative of \(f(x)\) on \([a, b]\). Then

\[ \int_a^b f(x)\, dx = F(b) - F(a). \]

1.2 Proof

Proof Idea (Part 1): Definition of derivative:

\[ A'(x) = \lim_{h \to 0} \frac{A(x+h) - A(x)}{h} \]

\[ A(x+h) - A(x) = \int_x^{x+h} f(t)\, dt \]

Since \(f\) is continuous at \(x\), for small \(h\) the function values \(f(t)\) are close to \(f(x)\). You can think of this as the area of a rectangle with width \(h\) and height approximately \(f(x)\).

\[ A'(x) = \lim_{h \to 0} \frac{f(x) \cdot h}{h} = f(x) \]

Hence, \(A'(x) = f(x)\).

Proof Idea (Part 2): Let \(F(x)\) be any antiderivative of \(f(x)\) on \([a,b]\).

\[ A(x) = \int_a^x f(t)\, dt \]

From Part 1, \(A'(x) = f(x)\), so \(A\) is an antiderivative of \(f\).

Any two antiderivatives differ by a constant. So there exists \(C\) such that:

\[ F(x) = A(x) + C \]

Evaluate at \(x = a\):

\[ F(a) = A(a) + C = 0 + C \implies C = F(a) \]

Thus,

\[ A(x) = F(x) - F(a) \]

Evaluate at \(x = b\):

\[ \int_a^b f(x)\, dx = A(b) = F(b) - F(a) \]

Hence, the Fundamental Theorem of Calculus (Part 2) is proved.

2. Examples

Example Problems for Part 1

Example P1.1: Derivative of an Integral

Compute

\[ \frac{d}{dx} \int_0^x \sin^2(t) \, dt \]

Solution: Let

\[ F(x) = \int_0^x \sin^2(t) \, dt. \]

By the Fundamental Theorem of Calculus, since \(\sin^2(t)\) is continuous, the derivative of \(F\) is simply the integrand evaluated at \(x\):

\[ F'(x) = \sin^2(x). \]

Therefore,

\[ \frac{d}{dx} \int_0^x \sin^2(t) \, dt = \sin^2(x). \]

Example P1.2: Derivative of an Integral with Variable Lower Limit

Compute

\[ \frac{d}{dx} \int_x^5 \sqrt{t^2 + 1} \, dt \]

Solution: By the Fundamental Theorem of Calculus (Part 1):

\[ \frac{d}{dx} \int_x^5 \sqrt{t^2 + 1} \, dt = \frac{d}{dx} \int_5^x - \sqrt{t^2 + 1} \, dt = -\sqrt{x^2 + 1}. \]

Example P1.3: Derivative of an Integral with a Function as Upper Limit

Compute \[ \frac{d}{dx} \int_1^{x^4} \sec(t) \, dt \]

Warning: It is NOT just \(\sec(x^4)\); we must account for the derivative of the upper limit.

Solution: Let us introduce a temporary variable:

\[ y = \int_1^{x^4} \sec(t) \, dt \]

Set \(u = x^4\). Then the integral becomes:

\[ y = \int_1^u \sec(t) \, dt \]

Differentiate using the chain rule:

\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

Apply the Fundamental Theorem of Calculus (Part 1) to the first factor. Explicitly,

\[ \frac{dy}{du} = \frac{d}{du} \int_1^u \sec(t) \, dt = \sec(u) \]

Substitute back \(u = x^4\) and differentiate \(u = x^4\):

\[ \frac{dy}{dx} = \sec(u) \cdot \frac{du}{dx} = \sec(x^4) \cdot 4x^3 = 4x^3 \sec(x^4) \]

Example Problems for Part 2

Example P2.1:

Compute \[ \int_1^3 e^x \, dx \] using the Fundamental Theorem of Calculus (Part 2).

Solution:

\[ F(x) = e^x \]

\[ \int_1^3 e^x \, dx = F(3) - F(1) \]

\[ \int_1^3 e^x \, dx = e^3 - e \]

Example P2.2:

\[ \int_{-1}^3 \frac{1}{x^2} \, dx \]

In order to use the Fundamental Theorem of Calculus, the function \(f(x)\) has to be defined and continuous for the whole interval \([a,b]\). Here it is not defined at \(0 \in [-1,3]\) (and hence not continuous), so we would treat it as an improper integral using a limit. We will learn about this later.

Difficult Problem:

Find the value of \(b > -1\) that maximizes

\[ \int_{-1}^{b} x^2 (5 - x) \, dx \]

Solution: Let

\[ F(b) = \int_{-1}^{b} x^2 (5 - x) \, dx, \quad F'(b) = b^2 (5 - b) \]

Critical points: \(F'(b) = 0 \Rightarrow b = 0, 5\)

Sign of \(F'(b)\) and behavior of \(F(b)\)

Interval	\(-1 < b < 0\)	\(0 < b < 5\)	\(b > 5\)
\(F'(b)\)	\(+\)	\(+\)	\(-\)
Behavior of \(F(b)\)	increasing	increasing	decreasing

From the table, the maximum occurs at:

\[ \boxed{b = 5} \]