Problem 63 - Applied Calculus

Question

Suppose \(\int_{0}^{2} g(t) d t=5 .\) Calculate the following: (a) \(\int_{0}^{4} g(t / 2) d t\) (b) \(\int_{0}^{2} g(2-t) d t\)

Step-by-Step Solution

Answer

(a) 10, (b) 5.

Step 1: Transform the integral for (a)

To calculate \( \int_{0}^{4} g(t/2) \, dt \), we use substitution. Let \( u = t/2 \), hence \( du = (1/2) \, dt \) or \( dt = 2 \, du \). Also, change the limits of integration: when \( t = 0, \, u = 0 \) and when \( t = 4, \, u = 2 \). Substitute these into the integral to obtain \( 2 \int_{0}^{2} g(u) \, du \).

Step 2: Evaluate the transformed integral for (a)

Using the substitution from Step 1, we have \( \int_{0}^{4} g(t/2) \, dt = 2 \int_{0}^{2} g(u) \, du \). We know from the given information that \( \int_{0}^{2} g(t) \, dt = 5 \). Therefore, \( 2 \int_{0}^{2} g(u) \, du = 2 \times 5 = 10 \).

Step 3: Transform the integral for (b)

For the integral \( \int_{0}^{2} g(2-t) \, dt \), we need to use substitution again. Let \( u = 2-t \), then \( du = -dt \). Changing the limits of integration: when \( t = 0, \, u = 2 \) and when \( t = 2, \, u = 0 \). This changes the integral to \( -\int_{2}^{0} g(u) \, du \), which is equal to \( \int_{0}^{2} g(u) \, du \) by reversing the limits and changing the sign.

Step 4: Evaluate the transformed integral for (b)

From the substitution made in Step 3, we have \( \int_{0}^{2} g(2-t) \, dt = \int_{0}^{2} g(u) \, du \), which is the same as \( \int_{0}^{2} g(t) \, dt \) given in the problem. Hence, the value is \( 5 \).

Key Concepts

Integration by Substitution

Integration by substitution is a method used to simplify the process of finding integrals. It is akin to reversing the chain rule for derivatives and can be exceptionally useful when dealing with composite functions.

When performing integration by substitution, the goal is to replace a complicated portion of the integral with a simpler variable. In our original example, to solve \(\int_{0}^{4} g(t/2) \, dt\), we used the substitution \(u = t/2\).

The substitution involves three main steps:

• Change the variable of integration from \(t\) to \(u\). This involves setting \(dt\) in terms of \(du\). For our example, \(du = (1/2) \, dt\), so \(dt = 2 \, du\).
• Change the limits of integration based on the new variable \(u\). Here, when \(t = 0\), \(u = 0\), and when \(t = 4\), \(u = 2\).
• Finally, insert the new limits and variable substitution into the integral, resulting in \(2 \int_{0}^{2} g(u) \, du\).

This simplifies the function, allowing us to integrate more easily. After solving the integral with the new limits and variable, simply revert back to the original variable if necessary.

Definite Integrals

Definite integrals calculate the net area under a curve within a specific interval. Unlike indefinite integrals, definite integrals provide a numeric value rather than a general function.

A definite integral is represented as \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the bounds or limits of integration, and \(f(x)\) is the function being integrated. This concept was central to the original problem, where we evaluated the integrals over specific intervals.

Steps to solve a definite integral include:

• Identifying the function \(f(x)\) and the interval \([a, b]\).
• Finding the antiderivative \(F(x)\) of \(f(x)\).
• Applying the Fundamental Theorem of Calculus, which states that the integral is \(F(b) - F(a)\).

In the exercise provided, the definite integral \(\int_{0}^{2} g(t) \, dt\) was known to be \(5\), making it straightforward to apply substitution methods to solve for related integrals.

Integral Transformations

Integral transformations allow us to change the form of an integral to enhance its evaluation or manipulation. These transformations often involve changing the variable of integration, limits, or the integrative approach to simplify the process.

In our example, we used transformations twice:

• First, in \(\int_{0}^{4} g(t/2) \, dt\), we transformed it to \(2 \int_{0}^{2} g(u) \, du\) by substituting \(u = t/2\).
• Second, in \(\int_{0}^{2} g(2-t) \, dt\), we transformed it using \(u = 2 - t\), which changed the integral to \(-\int_{2}^{0} g(u) \, du\). Reversing the limits, this becomes \(\int_{0}^{2} g(u) \, du\).

Transformations often make seemingly complex integrals more approachable and can provide insight into relationships between different functions or forms of functions. In practical terms, these transformations enable the application of known integrals to new problems, facilitating both problem-solving and understanding.