Problem 49 - Calculus

Question

Use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{0}^{1} \cos (3 x-3) d x $$

Step-by-Step Solution

Answer

\( \frac{1}{3} \sin(3) \)

Step 1: Identify the substitution

In the integral \(\int_{0}^{1} \cos (3x-3) \, dx\), we notice the composite function \(3x-3\) inside the cosine. We set the substitution \(u = 3x - 3\). This simplifies the integral by changing the variable.

Step 2: Compute the derivative

Differentiate the substitution \(u = 3x - 3\) to find \(du\). We have \(du/dx = 3\), therefore \(du = 3 \, dx\) or equivalently, \(dx = du/3\).

Step 3: Change the limits of integration

When \(x = 0\), the substitution gives \(u = 3(0) - 3 = -3\). When \(x = 1\), the substitution gives \(u = 3(1) - 3 = 0\). So, the limits of integration change accordingly, from \(x = 0\) to \(x = 1\) we have \(u = -3\) to \(u = 0\).

Step 4: Substitute and transform the integral

Substitute into the integral: \( \int_{0}^{1} \cos (3x-3) \, dx = \int_{-3}^{0} \cos u \, \frac{du}{3} \). This simplifies to \( \frac{1}{3} \int_{-3}^{0} \cos u \, du\).

Step 5: Evaluate the integral

The integral \( \frac{1}{3} \int_{-3}^{0} \cos u \, du\) can be computed using the antiderivative of \( \cos u\), which is \( \sin u\). So, \[ \frac{1}{3} (\sin u) \Big|_{-3}^{0} = \frac{1}{3} [\sin(0) - \sin(-3)] \].

Step 6: Simplify and compute the result

Evaluate \(\sin(0) - \sin(-3)\). Since \(\sin(0) = 0\) and \(\sin(-3) = -\sin(3)\), we have \(0 - (-\sin(3)) = \sin(3)\). So the integral evaluates to \( \frac{1}{3} \sin(3) \).

Key Concepts

Composite Function

A composite function occurs when you have one function inside of another. In the given exercise, we see this phenomenon with the term \(3x - 3\) inside the cosine function, \(\cos(3x - 3)\). The role of the composite function here is critical for applying the substitution rule.

• This composite nature allows us to simplify problems by changing variables using substitution.
• For substitution, we typically denote the inner function as \(u\), so in this case, \(u = 3x - 3\).
• Transforming the integral using \(u\) helps in adjusting the limits of integration and recalculating with simpler expressions.

Composite functions can be tricky at first, but they are incredibly useful in calculus for simplifying integral operations and calculations.

Antiderivative

The concept of an antiderivative is crucial when evaluating definite integrals, as it represents the function whose derivative is given. In the case of our integral \(\int_{-3}^{0} \cos u \, du\), the antiderivative of \(\cos u\) is \(\sin u\).

• Finding the antiderivative is essentially reversing the process of differentiation.
• Here, computing the antiderivative \(\sin u\) allows us to evaluate the integral more directly by considering the function values at given limits.
• This translates the problem into computing \[ \frac{1}{3} \cdot (\sin(u)) \big|_{-3}^{0} \], using simple trigonometric identities.

Remember, recognizing standard antiderivatives is a great skill, as it speeds up working with integrals tremendously!

Limits of Integration

Limits of integration define the boundary values for a definite integral, indicating where the integral starts and ends on the x-axis. In our transformation from \(x\) to \(u\), these limits heavily influence the final result.

• Originally, the limits are \(0\) to \(1\) for \(x\), corresponding to \(x = 0\) and \(x = 1\).
• After substitution, transform them to \(u = -3\) to \(u = 0\), following \(u = 3x - 3\).
• It's pivotal to accurately change these limits alongside the variable substitution so our evaluation stays consistent.

Mistaking these can lead to incorrect answers, so handling limits of integration is a careful process that aligns the substitution and integration steps together for accurate results.