Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \int\left(x^{3}+\sqrt{x}\right) d x $$

Step-by-Step Solution

Answer

\( \frac{x^{4}}{4} + \frac{2}{3}x^{3/2} + C \)

Step 1: Identify the Terms to Integrate

The integral given is \( \int\left(x^{3}+\sqrt{x}\right) d x \). We will break it into two parts: \( \int x^3 \, dx \) and \( \int \sqrt{x} \, dx \).

Step 2: Integrate the First Term

For the first term, \( \int x^3 \, dx \), we use the power rule for integration: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \). Thus, \( \int x^3 \, dx = \frac{x^{4}}{4} \).

Step 3: Integrate the Second Term

The second term is \( \int \sqrt{x} \, dx \), which can be rewritten as \( \int x^{1/2} \, dx \). Using the same power rule, \( \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} \). Simplifying gives \( \frac{2}{3}x^{3/2} \).

Step 4: Combine the Results

Combine the integrals of the two terms to obtain the final result: \( \int\left(x^{3}+\sqrt{x}\right) d x = \frac{x^{4}}{4} + \frac{2}{3}x^{3/2} + C \), where \( C \) is the constant of integration.

Key Concepts

Power Rule for Integration

The power rule for integration is a fundamental tool used in integral calculus for finding antiderivatives of power functions. It's a straightforward formula and is particularly helpful with polynomials. When you have a function of the form \( x^n \), the power rule states that the indefinite integral is:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

Here, \( C \) is the constant of integration, accounting for any constant term that was lost when differentiating. The only exception is when \( n = -1 \), since this would involve dividing by zero. In such cases, the natural log function, \( \ln|x| \), is the antiderivative instead.
In our exercise, we applied the power rule to two terms. For \( x^3 \), increasing the exponent by one yields \( x^4 \), and dividing by four gives us \( \frac{x^4}{4} \). Similarly, \( \sqrt{x} \) can be rewritten as \( x^{1/2} \), and the same process gives \( \frac{2}{3}x^{3/2} \).
Overall, understanding and mastering the power rule allows for finding the antiderivatives of polynomial terms quickly and easily.

Integration by Parts

The technique known as "integration by parts" is not directly applied in the given exercise. However, this method is vital for handling integrals where the power rule is insufficient. Integration by parts is derived from the product rule for differentiation. If you have a product of two functions, \( u(x) \) and \( v(x) \), the formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)

This technique is particularly useful when integrating the product of polynomial and exponential functions or logarithmic functions. The right choice of \( u \) and \( dv \) can simplify the computation significantly.
Applying integration by parts requires practice in choosing the correct functions to differentiate and integrate, but once mastered, it is an indispensable tool in a calculus student's toolkit.

Definite Integrals

While the exercise itself deals with indefinite integrals, it's essential to understand the concept of definite integrals as well. Definite integrals calculate the area under a curve between two specific points, \( a \) and \( b \), on the x-axis. This concept is formalized in the Fundamental Theorem of Calculus, which links the concept of a derivative with that of an integral.

• \( \int_a^b f(x) \, dx = F(b) - F(a) \)

Here, \( F \) is the antiderivative of \( f \). Evaluating \( F \) at the upper limit \( b \) and the lower limit \( a \), and finding their difference gives the net area.
Definite integrals specify a numerical result, as opposed to indefinite integrals, which contain a constant of integration \( C \) and represent a family of functions. Understanding definite integrals is key when solving real-world problems involving accumulation and total quantities.

Antiderivatives

An antiderivative is essentially the reverse process of differentiation. If you have a function \( f(x) \), an antiderivative of \( f \) is a function \( F(x) \) such that \( F'(x) = f(x) \). Integrating a function means finding all possible antiderivatives.

• A simple example is: if \( f(x) = x^2 \), then an antiderivative is \( F(x) = \frac{x^3}{3} + C \).

The integral sign \( \int \) indicates the process of finding an antiderivative, where each possible solution differs by a constant \( C \), due to the properties of derivatives (constants vanish during differentiation).
When dealing with antiderivatives, one often uses basic integration techniques, such as the power rule or substitution, to find these expressions. The constant of integration \( C \) is important as it represents the infinite set of all functions that have the derivative \( f(x) \). Understanding antiderivatives is fundamental to solving calculus problems and connecting derivatives to integrals.