Question

use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{-1}^{0} \sqrt{x^{3}+1}\left(3 x^{2}\right) d x $$

Short answer

The definite integral evaluates to \( \frac{2}{3} \).

Step-by-step solution

Identify the inner function and its derivative

The integral given is \( \int_{-1}^{0} \sqrt{x^{3}+1}(3x^{2}) \, dx \). First, identify the inner function in the integrand. Here, \( u = x^3 + 1 \). Calculate the derivative of \( u \) with respect to \( x \): \( \frac{du}{dx} = 3x^2 \). This matches the \( 3x^2 \, dx \) in the integrand.

Substitute using the substitution

Use the substitution \( u = x^3 + 1 \). Then \( du = 3x^2 \, dx \). Therefore, the original integral becomes \( \int \sqrt{u} \, du \).

Change the limits of integration

The original limits for \( x \) were from \( -1 \) to \( 0 \). Substitute these into \( u = x^3 + 1 \) to find the new limits for \( u \). When \( x = -1 \), \( u = (-1)^3 + 1 = 0 \). When \( x = 0 \), \( u = 0^3 + 1 = 1 \). Thus, the limits for \( u \) are \( 0 \) to \( 1 \).

Integrate with respect to \( u \)

The integral to be solved is \( \int_{0}^{1} \sqrt{u} \, du \). Use the power rule to integrate: \( \int u^{1/2} \, du = \frac{2}{3} u^{3/2} \).

Evaluate the definite integral

Apply the evaluation from \( u = 0 \) to \( u = 1 \): \[ \frac{2}{3} [u]_{0}^{1} = \frac{2}{3} (1^{3/2}) - \frac{2}{3} (0^{3/2}) = \frac{2}{3} \cdot 1 - 0 = \frac{2}{3} \].

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with the concept of integrals. It is the inverse process of differentiation and is used to find areas, volumes, central points, and many other useful things. The main idea is to sum up infinite small pieces to find a total. For example, think of calculating the area under a curve. Integrals can be used to determine that area. This process is about breaking down a large problem into infinitely small parts and then adding them up.
There are two major types of integrals in calculus:

• Indefinite integrals: These do not have limits of integration and result in a family of functions plus a constant (the constant of integration).
• Definite integrals: These have specified limits of integration and calculate a specific numerical value.

The exercise given is an example of how definite integrals are used in integral calculus to find exact quantities.

Definite Integrals

Definite integrals are a powerful tool in calculus, allowing us to compute the exact area under a curve between two points on the x-axis. Unlike indefinite integrals, which yield a general form of the antiderivative, definite integrals produce a specific value. This value can represent many things, such as total distance traveled or the net change of a quantity.
When working with definite integrals, you will often encounter:

• Limits of Integration: These are the bounds within which you are calculating the area. In the original exercise, the limits are from \(-1\) to \0\ for the variable \(x\), which change to \0\ to \1\ for \(u\) after substitution.
• Evaluation: Once the integral is computed, apply the limits of integration to obtain the final result. You substitute the upper limit and the lower limit into the antiderivative and then subtract the two results.

In the given exercise, converting the integral to terms of \(u\) simplifies the process and helps in neatly evaluating the integral within the new limits.

U-Substitution Method

The u-substitution method is a technique for evaluating integrals, especially useful when dealing with complex functions. It involves replacing a part of the integrand (the expression inside the integral) with a single variable, often denoted as \(u\), along with its differential \(du\). This substitution simplifies the integral, making it easier to solve.
The process often involves several key steps:

• Identify the Inner Function: Look for a function within the integrand whose derivative is also present in the integral. In our exercise, \(u = x^3 + 1\) simplifies the expression, with \frac{du}{dx} = 3x^2\.
• Perform the Substitution: Express the integral in terms of \(u\) and \(du\). In this case, the given integral is transformed into \int \sqrt{u} \, du\.
• Adjust the Limits: Change the original limits to correspond with the \(u\) variable. For \(x\) limits from \(-1\) to \0\, the \(u\) limits become \0\ to \1\.
• Integrate: Solve the new integral and evaluate it using the revised limits.

This method transforms the problem into a simpler form, allowing for straightforward integration and yielding precise results.