Problem 6 - Thomas Calculus

Question

Use the Substitution Formula in Theorem 6 to evaluate the integrals in Exercises \(1-24 .\) $$ \text { a. }\int_{0}^{\sqrt{7}} t\left(t^{2}+1\right)^{1 / 3} d t \quad \text { b. } \int_{-\sqrt{7}}^{0} t\left(t^{2}+1\right)^{1 / 3} d t $$

Step-by-Step Solution

Answer

Both integrals evaluate to \( \frac{3}{8} \times 7^{4/3} \).

Step 1: Identify the Substitution

For the integral \( \int t(t^2 + 1)^{1/3} \, dt \), set \( u = t^2 + 1 \). This means \( du = 2t \, dt \) or \( t \, dt = \frac{1}{2} du \). Substitute these into the integral.

Step 2: Rewrite the Integral with Substitution

Replacing \( t(t^2 + 1)^{1/3} \, dt \) with \( (u-1)^{1/3} \cdot \frac{1}{2} \, du \), rewrite the integral as \( \frac{1}{2} \int (u-1)^{1/3} \, du \).

Step 3: Change the Limits of Integration

For part (a), when \( t = 0 \), \( u = 1 \), and when \( t = \sqrt{7} \), \( u = 8 \). For part (b), when \( t = -\sqrt{7} \), \( u = 8 \), and when \( t = 0 \), \( u = 1 \).

Step 4: Evaluate the Integral

The integral becomes \( \frac{1}{2} \int (u - 1)^{1/3} \, du \). Integrate using the power rule: \( \int u^n \, du = \frac{u^{n+1}}{n+1} \). Thus, \( \frac{1}{2} \int (u-1)^{1/3} \, du \) evaluates to \( \frac{1}{2} \cdot \frac{3}{4} (u-1)^{4/3} + C \).

Step 5: Apply New Limits of Integration and Simplify

For part (a), evaluate \( \frac{3}{8} [(8-1)^{4/3} - (1-1)^{4/3}] = \frac{3}{8} [7^{4/3} - 0] \). For part (b), the evaluation is similar but reversed: \([-[(1-1)^{4/3}] + (8-1)^{4/3}]\), also simplified to \( \frac{3}{8} \times 7^{4/3} \).

Step 6: Combine Results and Interpret

Both parts (a) and (b) result in the same value as definite integrals reflect the symmetry in the bounds and function. The total integral value is the same.

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, representing the signed area under a curve between two points on the x-axis. These integrals are denoted by the limits of integration, which specify the starting and ending points. Essentially, it allows us to sum up infinitely small areas to find the total area.

• The process of evaluating a definite integral involves not just finding an antiderivative, but also subtracting values of this antiderivative at the upper and lower limits.
• This provides an exact numerical result rather than a general formula as in indefinite integration.

In our exercise, both parts (a) and (b) relied on definite integrals to evaluate the area under the given function. Despite having different initial bounds, they end up with symmetrical results because of the symmetry in the function and integration limits.

Integration Techniques

When faced with the task of finding an antiderivative, various integration techniques might be used. One powerful technique is substitution, especially for integrals that appear complex at first glance. Substitution involves replacing a segment of the integrand with a new variable to simplify integration.

• We identify a part of the integrand that can be substituted with a single variable, often simplifying the expression significantly.
• It's akin to reversing the chain rule within differentiation, allowing the integral to transform into a form that's easier to solve.

In the original solution, substitution simplified the given integrals. By letting \( u = t^2 + 1 \), the integral reduced to a simpler form where integration was manageable using the power rule.

Limits of Integration

The concept of the limits of integration is essential when dealing with definite integrals. These limits denote the interval over which you calculate the integral.

• The lower limit represents where the area under the curve begins, and the upper limit indicates where it ends.
• In substitution method, it's crucial to adjust these limits according to the new variable, ensuring they correspond correctly after substitution.

In this exercise, substituting \( t = 0 \) and \( t = \sqrt{7} \) changed the limits to \( u = 1 \) and \( u = 8 \) respectively for part (a). Similarly, different starting points were recalculated for part (b). Adjusting these was a key part of solving the integral correctly.

Power Rule for Integration

The power rule for integration is one of the most straightforward and commonly used techniques in calculus. The rule states that to integrate functions of the form \( x^n \), you can use the formula \( \int x^n \ dx = \frac{x^{n+1}}{n+1} + C \), provided \( n eq -1 \).

• This rule is especially useful when dealing with polynomial expressions after substitution has simplified the integrand.
• It helps convert the integral into a direct form that can be evaluated easily.

In the problem's solution, once the integration variable was changed to \( u \), the power rule was employed to find the antiderivative of the simplified expression \( (u-1)^{1/3} \). This approach led directly to an answer that was then evaluated over the specified limits, demonstrating the power rule's simplicity and utility.