Problem 190 - Calculus Volume 1

Question

In the following exercises, use the evaluation theorem to express the integral as a function \(F(x)\) . $$ \int_{a}^{x} t^{2} d t $$

Step-by-Step Solution

Answer

The integral is expressed as \( F(x) = \frac{x^3}{3} - \frac{a^3}{3} \).

Step 1: Understand the Problem

The given problem requires us to express the definite integral \( \int_{a}^{x} t^{2} \, dt \) as a function \( F(x) \) using the evaluation theorem, which is essentially the Fundamental Theorem of Calculus.

Step 2: Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( f \) is continuous on \([a, b]\), and \( F \) is an antiderivative of \( f \) on \([a, b]\), then \( \int_{a}^{x} f(t) \, dt = F(x) - F(a) \). In this case, \( f(t) = t^2 \).

Step 3: Find the Antiderivative

We need the antiderivative of \( t^2 \). The antiderivative of \( t^2 \) with respect to \( t \) is \( F(t) = \frac{t^3}{3} \).

Step 4: Apply the Evaluation Theorem

Substitute the antiderivative into the fundamental theorem: \( \int_{a}^{x} t^{2} \, dt = \left. \frac{t^3}{3} \right|_{a}^{x} = \frac{x^3}{3} - \frac{a^3}{3} \).

Step 5: Express as a Function

Thus, the integral expressed as a function \( F(x) \) is \( F(x) = \frac{x^3}{3} - \frac{a^3}{3} \).

Key Concepts

Definite Integral

A definite integral represents the signed area under a curve for a given interval. It gives us a numerical value that reflects this area, allowing us to summarize functions over specified intervals. In our exercise, we compute the integral from a specific starting point, \(a\), to an endpoint, \(x\), of the function \( t^2 \).
This process involves summing all infinitesimally small pieces of the function's curve between these two points.

• Definite integrals have limits, \(a\) and \(x\), defining the interval over which the integration occurs.
• The result of a definite integral is always a number that represents the net area.

This can be crucial in fields like physics and engineering where understanding the cumulative effect or total change over an interval is necessary. In simple terms, think of it as adding up many tiny changes to get a big picture result.

Antiderivative

An antiderivative, also known as an indefinite integral, reverses the process of differentiation. It finds a function whose derivative equals the original function we started with. For the function \( t^2 \), its antiderivative is found by searching for a function \( F(t) \) that satisfies \( F'(t) = t^2 \).
Using a basic rule of antiderivatives:\[F(t) = \frac{t^3}{3}\]This antiderivative makes sure that when differentiated, we return to our original function \( t^2 \).

• When finding an antiderivative, a constant \( C \) is added since derivatives of constants are zero.
• The antiderivative of \( t^2 \) is particularly \( \frac{t^3}{3} + C \).

This step is vital in applying the Fundamental Theorem of Calculus, which requires an antiderivative to evaluate a definite integral.

Evaluation Theorem

The evaluation theorem is an alternative name for the Fundamental Theorem of Calculus, which connects differentiation and integration. It acts as a bridge between the antiderivative of a function and its definite integral.
To apply this theorem, you need to:

• First, find an antiderivative \( F(t) \) of the function \( f(t) \).
• Evaluate \( F(t) \) at the upper limit \( x \) and subtract the evaluation at the lower limit \( a \).

The theorem states:
\[\int_{a}^{x} t^2 \, dt = F(x) - F(a)\]With the antiderivative \( \frac{t^3}{3} \), we compute:
\[\int_{a}^{x} t^2 \, dt = \left. \frac{t^3}{3} \right|_a^x = \frac{x^3}{3} - \frac{a^3}{3}\]In essence, this theorem simplifies the complex task of calculating areas and makes understanding definite integrals straightforward by reducing it to basic arithmetic using antiderivatives.