Problem 68 - Calculus Early Transcendentals

Question

Miscellaneous integrals Evaluate the following integrals. \(\int_{0}^{1} \frac{16^{x}}{4^{2 x}} d x\)

Step-by-Step Solution

Answer

Based on the given step by step solution, the integral \(\int_{0}^{1} \frac{16^{x}}{4^{2 x}} d x\) evaluates to 1.

Step 1: Simplify the integrand

Before evaluating the integral, let's simplify the integrand. The expression can be written as: \(\int_{0}^{1} \frac{16^{x}}{4^{2 x}} d x = \int_{0}^{1} (\frac{16}{4^2})^{x} d x\) Now, we know that \(16 = 4^2\), so we can rewrite the expression as: \(\int_{0}^{1} (\frac{4^2}{4^2})^{x} d x = \int_{0}^{1} (1)^{x} d x\) Since \(1^x = 1\) for any real value of \(x\), the integral becomes: \(\int_{0}^{1} 1 d x\)

Step 2: Evaluate the integral

Now, let's evaluate the integral \(\int_{0}^{1} 1 d x \). Since the function is a constant (1), the antiderivative of 1 with respect to \(x\) is simply \(x\). Thus, \(\int_{0}^{1} 1 d x = x\Big|_0^1\) Now, apply the fundamental theorem of calculus: \((x\Big|_0^1) = (1 - 0) = 1\) So the integral value is 1. The integral \(\int_{0}^{1} \frac{16^{x}}{4^{2 x}} d x\) evaluates to 1.

Key Concepts

Simplify the Integrand

Simplifying the integrand is a powerful technique that often makes an integral much easier to solve. When looking at an expression like \( \frac{16^{x}}{4^{2x}} \), simplifying means rewriting it into a more manageable form.
The given expression \( \frac{16^{x}}{4^{2x}} \) can initially look complex, but we can simplify it by recognizing a common base:

• Observe that \(16 = 4^2\).
• Rewrite the integrand: \( \frac{16^{x}}{4^{2x}} = \left( \frac{4^{2}}{4^2} \right)^{x} \).
• This simplifies to \( (1)^x \), as \( \frac{4^{2x}}{4^{2x}} = 1 \).

With this simplification, the integral reduces to \( \int_{0}^{1} 1 \, dx \). This step makes the whole integration process more straightforward, setting the stage for easier computation.

Antiderivative

Finding the antiderivative is a fundamental skill in integration. An antiderivative of a function is another function whose derivative is the original function. For instance, when dealing with a constant function such as \( 1 \), its antiderivative is simply \( x \), since differentiating \( x \) gives \( 1 \).
Therefore, when we look at the integral \( \int_{0}^{1} 1 \, dx \), we compute:

• The antiderivative of \( 1 \) in respect to \( x \) is \( x \).
• Hence, \( \int 1 \, dx = x + C \), where \( C \) is the constant of integration that is omitted when evaluating definite integrals.

This antiderivative allows us to apply the bounds \( x \Big|_0^1 \) later on, to compute the definite integral.

Fundamental Theorem of Calculus

The fundamental theorem of calculus connects differentiation with integration, providing an efficient way to evaluate definite integrals. It essentially states that if you have an antiderivative \( F(x) \) of a function \( f(x) \), the definite integral of \( f(x) \) from \( a \) to \( b \) is \( F(b) - F(a) \).
In our case, after obtaining the antiderivative \( x \) for the function \( 1 \), we use the theorem to evaluate:

• Apply the limits of integration: \( x \Big|_0^1 \).
• Calculate as \( 1 - 0 \), resulting in \( 1 \).

This result, \( 1 \), gives us the value of the original definite integral \( \int_{0}^{1} \frac{16^{x}}{4^{2x}} \, dx \). By leveraging the fundamental theorem, we ensure that complex processes, like integrating, can be completed accurately and efficiently.