Problem 29 - Calculus

Question

Find \(\int_{1}^{x} v(t) d t\) from the facts. $$ \int_{0}^{x} v(t) d t=\frac{x}{x+2}. $$

Step-by-Step Solution

Answer

\( \int_{1}^{x} v(t) \, dt = \frac{x}{x+2} - \frac{1}{3} \).

Step 1: Understand the Problem

We are given the function \( v(t) \) such that if we integrate from 0 to \( x \), we get \( \int_{0}^{x} v(t) \, dt = \frac{x}{x+2} \). We need to find \( \int_{1}^{x} v(t) \, dt \).

Step 2: Apply the Property of Definite Integrals

We know from the properties of definite integrals that \( \int_{0}^{x} v(t) \, dt = \int_{0}^{1} v(t) \, dt + \int_{1}^{x} v(t) \, dt \). Our goal is to find \( \int_{1}^{x} v(t) \, dt \).

Step 3: Rearrange to Solve for the Desired Integral

Rearrange the equation: \( \int_{1}^{x} v(t) \, dt = \int_{0}^{x} v(t) \, dt - \int_{0}^{1} v(t) \, dt \). Since we know \( \int_{0}^{x} v(t) \, dt = \frac{x}{x+2} \), we need to find \( \int_{0}^{1} v(t) \, dt \).

Step 4: Evaluate \( \int_{0}^{1} v(t) \, dt \) Using Given Information

Plug in \( x = 1 \) in the given function: \( \int_{0}^{1} v(t) \, dt = \frac{1}{1+2} = \frac{1}{3} \).

Step 5: Substitute and Simplify

Substitute back into the rearranged equation: \( \int_{1}^{x} v(t) \, dt = \frac{x}{x+2} - \frac{1}{3} \). This is the required integral of \( v(t) \) from 1 to \( x \).

Key Concepts

Properties of Integrals

Understanding the properties of integrals is crucial when solving problems involving definite integrals. One of the most powerful properties to remember is the linearity and additivity of integrals. These properties allow us to break down complex integrals into simpler parts that can be easily managed.
For definite integrals, one key property is:

• If you have a definite integral from a to b, it can be split into two parts: \[ \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx = \int_{a}^{b} f(x) \, dx \]

This property is particularly useful when our goal is to compute an integral over a specific range but only the integral over a larger range is initially known.
In our exercise, this property helps us express \( \int_{1}^{x} v(t) \, dt \) in terms of other known integrals, particularly knowing \( \int_{0}^{x} v(t) \, dt \). By rearranging and using the additivity, we achieve the expression we need.

Integration Techniques

Integration techniques revolve around finding effective methods to solve integrals, whether definite or indefinite. One common technique involves substitution, where we replace variables or expressions to simplify the integral.
In problems like our example, substitution is not directly used, but we manipulate known integrals to find the desired solution. The problem involved understanding and re-arranging the integral from 0 to \( x \), using known outcomes of integration. Such techniques involve logical reasoning and transformation of expressions.
It's essential to recognize when to apply specific properties like parts, substitution, or recognizing patterns. In our problem, the focus was on re-arranging and using given conditions straightforwardly, which is a simplified demonstration of applying properties rather than computation-heavy techniques.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, providing a practical way to evaluate definite integrals. It states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then:

• The integral of \( f \) over \([a, b]\) can be expressed as \( F(b) - F(a) \).

For the current exercise, while the theorem itself is not directly invoked as it does not deal with finding an antiderivative explicitly, the foundational understanding aids us when considering how to evaluate and manipulate integrals like \( \int_{1}^{x} v(t) \, dt \).
The theorem also justifies and underpins the rearrangement of integrals and using specific values as done when \( x = 1 \) was substituted into the equation. The crucial understanding is that integration sums the area under the curve, which is effectively captured when we split integrals using properties like those in earlier sections.