Question

use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{-3}^{3} \sqrt{7+2 t^{2}}(8 t) d t $$

Short answer

The definite integral evaluates to 0.

Step-by-step solution

Choose a Substitution

Identify a substitution that simplifies the integrand. Let \( u = 7 + 2t^2 \). Then, the derivative \( du = 4t \, dt \). Rearrange this to find \( dt \): \( dt = \frac{du}{4t} \).

Adjust the Integral for Substitution

Replace \( \sqrt{7+2t^2} \) with \( \sqrt{u} \) and \( 8t \, dt \) with \( 2du \) (since \( 8t \, dt = 2 \cdot 4t \, dt = 2du \)). The integral becomes \( \int \sqrt{u} \, 2 \, du \).

Change Limits of Integration

Calculate the new limits of integration. When \( t = -3 \), \( u = 7 + 2(-3)^2 = 25 \). When \( t = 3 \), \( u = 7 + 2(3)^2 = 25 \). Thus, the limits of integration remain \( 25 \) to \( 25 \).

Evaluate the Integral with New Limits

Since the limits of integration are the same (from 25 to 25), the integral becomes zero, i.e., \( \int_{25}^{25} \sqrt{u} \, 2 \, du = 0 \).

Conclude the Result

The integral evaluates to 0 because the upper and lower limits of integration are equal, meaning the area under the curve is zero.

Key concepts

Substitution Rule

The Substitution Rule is a powerful technique in calculus that allows us to simplify complex integrals. By changing variables, we can transform an intimidating expression into something more manageable. This is particularly useful when dealing with integrals involving composite functions. In essence, the substitution rule finds a new variable (often denoted as \( u \)) which replaces a compound part of the integrand.
This helps us in integrating functions that otherwise seem difficult to handle. When performing substitution, we must also change the differential \( dt \) to \( du \), using the derivative of the substitution function. Overall, substitution is akin to applying the chain rule in reverse to simplify the integration process.

Integral Calculus

Integral calculus is one-half of calculus, concentrating on accumulation or the sum total of quantities. In simpler terms, it helps find the total value given the rate of change. Definite integrals, a special category in integral calculus, allow us to compute the actual quantity given the bounds.
Integrals calculate areas under curves or accumulation of quantities, making them crucial for physics, engineering, and numerous fields. In this context, using the substitution rule within integral calculus can simplify the undertaking of evaluating definite integrals.

Limits of Integration

The limits of integration are crucial when dealing with definite integrals. These limits represent the bounds within which we sum or accumulate values. In definite integrals, these limits determine the range over which the integrand is applied.
For instance, in our original exercise, the limits were \(-3\) to \(3\) but changed to \(25\) to \(25\) upon substitution. When the upper and lower limits are equal, the accumulated area between them is zero. This highlights the importance of paying attention to these boundaries when evaluating definite integrals.

Area under the Curve

One of the most common applications of definite integrals is finding the area under a curve. The integral gathers an infinite number of infinitesimally small areas under a function's curve to give a total area. Conceptually, if you plot the function on a graph and want to know the area it covers between two points on the x-axis, definite integrals come into play.
In scenarios where the limits are equal, such as in our exercise after substitution, the area under the curve is zero because there is no distance covered horizontally. Integrals give a geometric interpretation, translating between equations and visual areas, crucial in numerous real-world scenarios.