Problem 47 - Calculus

Question

Use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{0}^{\pi / 6} \sin ^{3} \theta \cos \theta d \theta $$

Step-by-Step Solution

Answer

The definite integral is \(\frac{1}{16}\).

Step 1: Identify the Substitution

To simplify the integral, we perform a substitution. Notice the function \(\sin ^{3} \theta \, \cos \theta\). If we let \( u = \sin\theta\), then \( d\theta = \frac{du}{\cos\theta}\). Thus, \( \cos \theta \, d\theta = du\).

Step 2: Change the Limits of Integration

With substitution \(u = \sin\theta\), we need to change the limits of integration. When \(\theta = 0\), \(u = \sin(0) = 0\). When \(\theta = \frac{\pi}{6}\), \(u = \sin(\frac{\pi}{6}) = \frac{1}{2}\). So, the new limits are from \(u = 0\) to \(u = \frac{1}{2}\).

Step 3: Rewrite the Integral

Substituting back into the integral, we have: \(\int_{0}^{1/2} u^3 \, du\).Here, we replaced \(\sin^3 \theta \) by \(u^3\) and \(\cos \theta \, d\theta\) by \(du\).

Step 4: Evaluate the Integral

The integral \(\int u^3 \, du\) is straightforward to compute. It is equal to \(\frac{u^4}{4}\). So we evaluate this from \(0\) to \(\frac{1}{2}\): \[\left[\frac{u^4}{4}\right]_{0}^{1/2} = \frac{\left(\frac{1}{2}\right)^4}{4} - \frac{0^4}{4} = \frac{1}{16}.\]

Step 5: Simplify the Result

After evaluating the expression, the result of the original integral is \(\frac{1}{16}\), which is the area under the curve of \(\sin^3 \theta \cos \theta\) from \(\theta = 0\) to \(\theta = \frac{\pi}{6}\).

Key Concepts

Definite Integrals

Definite integrals are used to calculate the signed area under a curve across a particular interval. The integral symbol, together with upper and lower limits, indicates we are dealing with a definite integral. In this exercise, the function \( \sin^3 \theta \cos \theta \) is evaluated from \( \theta = 0 \) to \( \theta = \frac{\pi}{6} \).
To solve a definite integral, we use the antiderivative (or the indefinite integral) and evaluate it using the Fundamental Theorem of Calculus. This process involves determining the antiderivative of the function, then subtracting its value at the lower limit from its value at the upper limit.
For this specific integral, the function within the integral is transformed using a substitution to simplify the computation, yet the principle of computing the signed area remains the same.

Limits of Integration

The limits of integration in a definite integral define the range over which the integral is evaluated. In this case, the limits are from \( \theta = 0 \) to \( \theta = \frac{\pi}{6} \).
When using substitution, these limits must be adjusted according to the new variable of integration. Here, we substituted \( u = \sin \theta \), leading to new integration limits. When \( \theta = 0 \), \( u = \sin(0) = 0 \), and when \( \theta = \frac{\pi}{6} \), \( u = \sin(\frac{\pi}{6}) = \frac{1}{2} \). Therefore, the limits of integration change from \( \theta \) values to \( u \) values from 0 to \( \frac{1}{2} \).
Adjusting limits correctly is vital because it lets us fully transform the integral into the new variable, simplifying computation and ultimately allowing accurate evaluation of the integral.

Integral Evaluation

Evaluating the integral consists of computing its antiderivative and then applying the limits of integration. Once the function is transformed using substitution and its new limits are defined, we use these limits to execute the evaluation.
In our modified integral, \( \int_{0}^{1/2} u^3 \, du \), the antiderivative of \( u^3 \) is \( \frac{u^4}{4} \). After determining this, we apply the limits of integration from 0 to \( \frac{1}{2} \).

• Evaluate at the upper limit: \( \left(\frac{1}{2}\right)^4 = \frac{1}{16} \)
• Evaluate at the lower limit: \( 0^4 = 0 \)
• Subtract the two values: \( \frac{1}{16} - 0 = \frac{1}{16} \)

The result, \( \frac{1}{16} \), represents the exact computed area under the curve for the original definite integral, providing a complete evaluation of the initial problem.