Question

use the method of substitution to find each of the following indefinite integrals. $$ \int x \sqrt{x^{2}+4} d x $$

Short answer

\( \int x \sqrt{x^{2}+4} \, dx = \frac{1}{3} (x^{2}+4)^{3/2} + C \)

Step-by-step solution

Choose a Substitution

We need to choose an appropriate substitution for the given integral. Looking at the integral \( \int x \sqrt{x^2 + 4} \, dx \), it is beneficial to substitute the expression inside the square root. Let's set \( u = x^2 + 4 \). This implies that \( du = 2x \, dx \).

Solve for dx

Since the substitution gives us \( du = 2x \ dx \), we can solve for \( dx \) as follows: \( dx = \frac{du}{2x} \).

Substitute in the Integral

Substitute \( u = x^2 + 4 \) and \( dx = \frac{du}{2x} \) into the integral. This transforms the original integral:\[ \int x \sqrt{x^2+4} \, dx = \int x \sqrt{u} \, \frac{du}{2x} \]Simplify this to get:\[ \frac{1}{2} \int \sqrt{u} \, du \]

Integrate with Respect to u

Now we integrate \( \frac{1}{2} \int \sqrt{u} \, du \). Recalling that \( \sqrt{u} = u^{1/2} \), the integral becomes:\[ \frac{1}{2} \int u^{1/2} \, du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} = \frac{1}{3} u^{3/2} \]

Substitute Back to Original Variable

Finally, substitute back \( u = x^2 + 4 \) into the expression to return to the variable \( x \). The final result is:\[ \frac{1}{3} (x^2 + 4)^{3/2} + C \] where \( C \) is the constant of integration.

Key concepts

Indefinite Integrals

Indefinite integrals are an essential component of calculus that represent a family of functions. The term **_indefinite_** is used because the integration includes a constant of integration, which makes the solution collection non-specific in its value.

For instance, when we find the indefinite integral of a function, we aim to determine an antiderivative that can recreate the original function when differentiated. This results in:

• A general form that includes an arbitrary constant, often denoted as 'C', to express any vertically shifted version of the antiderivative.

In mathematical notation, if we have a function represented as \( F(x) \), the indefinite integral of its derivative \( f(x) \) is noted as: \[ \int f(x) \, dx = F(x) + C \] This principle of indefinite integrals lays the groundwork to evaluate and simplify more complex integral equations.

Substitution Method

The substitution method is a powerful technique in integral calculus, assisting in simplifying integrals that appear complex at first glance. In this approach, a substitution is made for a part of the integral, which will transform it into a format that is simpler to evaluate.

To effectively use substitution, follow these general steps:

• Identify a portion of the integrand that can be replaced with a simpler variable, such as \( u \).
• Compute the differential \( du \) of the substitution and express the initial integral's derivative \( dx \) in terms of \( du \).
• Substitute \( u \) and \( du \) into the integral to simplify it, often eliminating a complex relationship.
• Integrate with respect to \( u \).
• Finally, substitute the original variable back into the solution for the final integrating result.

For example, in the exercise \( \int x \sqrt{x^2 + 4} \, dx \), the substitution \( u = x^2 + 4 \) is made, transforming the problem to a simpler integral which can be easily solved.

Integral Calculus

Integral calculus forms one of the two principal areas of calculus, the other being differential calculus. It essentially revolves around the concepts of integration and antidifferentiation, which involve calculating the area under curves or recovering a function from its derivative.

Some foundational goals include:

• Determining the total accumulation of a quantity, which in geometric terms can mean the area under a curve described by a function.
• Using antiderivatives or indefinite integrals to reverse differentiation, thus reconstructing a function given its rate of change.

Integral calculus is crucial in understanding properties such as total change between two points. Beyond simple areas, it is the backbone for numerous applications:

• Calculating volumes, areas, and arc lengths.
• Solving differential equations that appear in physics, engineering, and other sciences.

By mastering both the fundamentals of indefinite integrals and methods like substitution, students can tackle a vast array of problems in integral calculus.