Problem 448 - Calculus Volume 1

Question

Evaluate the following integrals. $$\int_{0}^{4} \frac{3 t}{\sqrt{1+6 t^{2}}} d t$$

Step-by-Step Solution

Answer

The integral evaluates to approximately 18.67.

Step 1: Identify the method

The integral \( \int_{0}^{4} \frac{3t}{\sqrt{1+6t^2}} \, dt \) suggests the use of substitution, typically beneficial with square roots and products.

Step 2: Choose a substitution

Set \( u = 1 + 6t^2 \). Then \( du = 12t dt \), leading to \( \frac{du}{12} = t dt \).

Step 3: Change the limits

Modify the limits according to the substitution: When \( t = 0, u = 1 \) and when \( t = 4, u = 1 + 6(4^2) = 97 \).

Step 4: Substitute and rewrite the integral

Substitute \( t dt \) with \( \frac{du}{12} \) and \( 3t \) becomes \( \frac{u - 1}{2} \) from \( 1 + 6t^2 = u \):\[\frac{1}{4} \int_{1}^{97} \frac{1}{\sqrt{u}} \, du\]

Step 5: Simplify the integral expression

This integral simplifies to \(\frac{1}{4} \times 2 \times (u^{1/2}) + C\) evaluated from 1 to 97, giving:\[\left[ 2 \sqrt{u} \right]_{1}^{97}\]

Step 6: Evaluate the definite integral

Evaluate the expression: \[2 \sqrt{97} - 2 \sqrt{1} = 2(\sqrt{97} - 1)\]

Step 7: Calculate the numerical result

Calculate \( 2(\sqrt{97} - 1) \) using a calculator for exactness, which yields approximately \( 18.67 \).

Key Concepts

Definite Integral

The concept of a definite integral is an essential part of integral calculus, specifically used to compute the total accumulation of quantities, like area under a curve, over a specific interval. In the context of the exercise, we evaluate the definite integral from 0 to 4 of the given function. The definite integral is denoted by the integral symbol with upper and lower limits, precisely:

• The lower limit (0 in this case) represents the starting value of the independent variable, often time or another parameter.
• The upper limit (4 here) signifies the ending value.

When calculating a definite integral, the result is a specific numerical value, unlike an indefinite integral, which represents a family of functions. This value represents the total net area under the curve between the two specified points on the x-axis.

Substitution Method

The substitution method in integral calculus is a technique used to simplify complex integrals, making them easier to compute. In this scenario, substitution is particularly useful as the integral involves a product and a square root function.

The substitution step involves:

• Identifying a portion of the integrand (the expression inside the integral) that can be set as a new variable, often denoted as "u." For example, letting \( u = 1 + 6t^2 \) simplifies the expression under the square root.
• Calculating the differential of \( u \) which connects back to the original variable, here given by \( du = 12t \, dt \).
• Rewriting both the limits of integration and the integrand in terms of the new variable \( u \).

This substitution reduces the original expression into a simpler form that's easier to integrate.

Square Roots

Square roots often appear in integrals and can complicate the integration process because of their nonlinear properties. In this exercise, the square root \( \sqrt{1 + 6t^2} \) appears in the denominator, suggesting the need for an effective technique to handle integrations involving radicals.

Substituting \( u = 1 + 6t^2 \) helps us manage the square root seamlessly by transforming it in terms of \( u \). Thus, our integral transforms into a simpler one - \( \frac{1}{\sqrt{u}} \) - which is more straightforward to solve.

The goal is to reduce the complexity associated with the square root and facilitate a more direct evaluation.

Integration Limits

Integration limits are crucial in evaluating a definite integral. These set the boundaries for the interval over which the function is integrated. Initially, the limits are based on the original variable.

When using the substitution method, you must adjust the integration limits to correspond with the new variable \( u \):

• Find the value of \( u \) when the original variable is at the lower limit, here \( t = 0 \), resulting in \( u = 1 \).
• Determine the value of \( u \) when the original variable is at the upper limit, \( t = 4 \), giving \( u = 97 \).

This conversion is vital as it ensures that the integration is computed correctly over the new interval, reflecting the original interval in terms of \( u \). The transformed limits allow you to evaluate the integral entirely within the scope of the substitution.