Problem 55 - Calculus: Early Transcendentals

Question

Use the properties of integrals to verify the inequality without evaluating the integrals. \( \displaystyle \int^4_0 (x^2 - 4x + 4) \,dx \ge 0 \)

Step-by-Step Solution

Answer

The inequality holds because \( (x-2)^2 \) is non-negative for all \( x \) in the interval [0, 4].

Step 1: Analyze the quadratic function

The function inside the integral is \( f(x) = x^2 - 4x + 4 \). To understand its behavior, let's rewrite it in vertex form \( f(x) = (x-2)^2 \). This shows that \( f(x) \) is a perfect square and reaches a minimum value of 0 at \( x = 2 \).

Step 2: Consider properties of squares

Since \( (x-2)^2 \) is always non-negative (as squares of real numbers are always non-negative), the integrand \( x^2 - 4x + 4 \) is non-negative for all \( x \) in the interval \([0, 4]\).

Step 3: Apply properties of integrals

Integrals of non-negative functions over a closed interval are non-negative. Therefore, \( \int_{0}^{4} (x^2 - 4x + 4) \, dx \geq 0 \).

Step 4: Conclude that inequality holds

Since the integrand is non-negative for all \( x \) in the interval and based on the properties of integrals, we can conclude that the integral \( \int_{0}^{4} (x^2 - 4x + 4) \, dx \geq 0 \) without computing the integral.

Key Concepts

Quadratic Functions

Quadratic functions are an essential part of mathematics, often appearing in the form of polynomials with a degree of 2. A quadratic function is typically expressed as \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The graph of a quadratic function is a curve called a parabola, which can open upwards or downwards depending on the sign of \( a \). In the original problem, the quadratic function was \( x^2 - 4x + 4 \). This particular quadratic is called a perfect square trinomial, making it easy to factor into \( (x-2)^2 \). This form is crucial in understanding the behavior and properties of the function.

Non-negative Functions

A non-negative function is one that does not take on negative values over its domain. This characteristic is particularly significant when it comes to integration, as integrating a non-negative function over an interval will yield a non-negative result. In the context of the exercise, we considered the function \( (x-2)^2 \). Since squares of real numbers are always non-negative, \( (x-2)^2 \ge 0 \) for all values of \( x \). Thus, the function \( x^2 - 4x + 4 \) is non-negative for every point between 0 and 4, aligning with its behavior as a non-negative quadratic function.

Vertex Form

The vertex form of a quadratic function helps in quickly identifying some of the key properties, such as the vertex, which is a crucial point of the parabola. The vertex form is expressed as \( f(x) = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola. For the problem, converting \( x^2 - 4x + 4 \) into vertex form \( (x-2)^2 \) reveals that the vertex of the parabola occurs at \( x = 2 \), where it achieves a minimum value of 0. This conversion not only simplifies the examination of the function's range but also aids in recognizing its non-negative nature on the specified interval.

Integral Inequality

An integral inequality involves comparing the integral of a function over a given interval to other numeric values, typically proving one side is greater than or equal to the other. Often, this utilizes properties of the function or the integral to reach a conclusion without solving the integral explicitly. In the exercise, the inequality \( \int_{0}^{4}(x^2 - 4x + 4) \,dx \geq 0 \) is affirmed by recognizing the non-negative nature of the integrand across the interval \([0, 4]\). Since the function \((x-2)^2\) is always non-negative, the integral respects the inequality property, allowing us to confidently assert that the integral value is non-negative without direct computation.