Question

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

Short answer

\( \frac{1}{2}(\sqrt{97} - 1) \)

Step-by-step solution

Identify the Type of Integral

The integral to evaluate is \( \int_{0}^{4} \frac{3t}{\sqrt{1+6t^2}} \, dt \). This integral involves a square root in the denominator, suggesting a possible substitution method typically used for integrals of the form \( \int \frac{u}{\sqrt{a^2 + u^2}} \, du \).

Choose a Suitable Substitution

To simplify the integral, choose the substitution \( u = 1 + 6t^2 \). Then differentiate: \( du = 12t \, dt \), which implies \( t \, dt = \frac{1}{12} du \). Rewriting the integral, adjust the limits of integration: - When \( t = 0 \), \( u = 1 \). - When \( t = 4 \), \( u = 1 + 6(4)^2 = 97 \).

Rewrite the Integral under the New Substitution

Substitute and rewrite the integral in terms of \( u \): \[ \int_{1}^{97} \frac{3t}{\sqrt{u}} \times \frac{1}{12} du = \frac{1}{4} \int_{1}^{97} \frac{1}{\sqrt{u}} \, du. \]

Evaluate the Simplified Integral

The integral \( \int \frac{1}{\sqrt{u}} \, du \) is recognized as \( \int u^{-1/2} \, du \). The antiderivative is \( 2u^{1/2} \). Evaluate from 1 to 97:\[ \frac{1}{4} \times [2u^{1/2}]_{1}^{97} = \frac{1}{4} \times (2\sqrt{97} - 2\sqrt{1}). \]

Calculate the Final Answer

Continuing from the previous step: \[ \frac{1}{4} \times (2\sqrt{97} - 2) = \frac{1}{2}(\sqrt{97} - 1). \] Thus, the evaluated integral is \( \frac{1}{2}(\sqrt{97} - 1) \).

Key concepts

Substitution Method

When faced with complex integrals, the substitution method is a powerful tool. It involves replacing a part of the integral with a new variable, simplifying the function. In our exercise, we encounter the integral \( \int_{0}^{4} \frac{3t}{\sqrt{1+6t^2}} \, dt \).
This integral has a square root in the denominator. By choosing \( u = 1 + 6t^2 \), a part of the integrand simplifies.

• Differentiate our substitution: \( du = 12t \, dt \)
• Rearrange: \( t \, dt = \frac{1}{12} du \)

This simplification transforms the integral into something more manageable by reducing its complexity, making it easier to find the antiderivative.

Definite Integrals

Definite integrals are a way to calculate the net area under a curve, from a start point to an end point. In our original problem, we work with the limits from 0 to 4.
After performing the substitution \( u = 1 + 6t^2 \), we update these limits: as \( t \) varies, so do the values of \( u \).

• When \( t = 0 \), \( u = 1 \)
• When \( t = 4 \), \( u = 97 \)

Integrating within these new limits ensures the computed area reflects the original interval in terms of \( t \). Remembering to change limits is crucial when substituting, because it impacts the bounds of the integral's evaluation.

Integral Evaluation Methods

Choosing the right technique to evaluate integrals can make the difference between an easy solution and a daunting task. In this exercise, we use a combination of substitution and basic antiderivative rules.
Once substitution is done, we get the integral \( \frac{1}{4} \int_{1}^{97} \frac{1}{\sqrt{u}} \, du \). This is recognized as \( \int u^{-1/2} \, du \).

• The antiderivative is \( 2u^{1/2} \)
• Evaluate from 1 to 97: \( 2\sqrt{97} - 2\sqrt{1} \)

Finally, multiply through the constant \( \frac{1}{4} \) to get \( \frac{1}{2}(\sqrt{97} - 1) \). This method highlights how strategic substitution simplifies integration and points to how integral evaluation relies on identifying familiar forms.