Question

Evaluate the following integrals. Include absolute values only when needed. $$\int_{0}^{1} \frac{16^{x}}{4^{2 x}} d x$$

Short answer

Question: Calculate the definite integral of the given function: \(\int_{0}^{1} \frac{16^x}{4^{2x}} dx\) Answer: 1

Step-by-step solution

Simplify the integrand

We begin by simplifying the integrand \(\frac{16^x}{4^{2x}}\). Using the rule of exponents \(a^{mx} = (a^m)^x\), we can rewrite the integrand as follows: $$\frac{16^x}{4^{2x}} = \frac{(2^4)^x}{(2^2)^{2x}} = \frac{2^{4x}}{2^{4x}}.$$ Now, we can divide the powers of 2: $$\frac{2^{4x}}{2^{4x}} = 2^{4x-4x} = 2^0 = 1.$$ So the integrand simplifies to 1: $$\int_{0}^{1} \frac{16^x}{4^{2x}} dx = \int_{0}^{1} 1 dx.$$

Find the antiderivative

Next, we find the antiderivative of 1 with respect to x: $$\int 1 dx = x + C.$$

Evaluate the integral

Now, we can evaluate the integral using the bounds 0 and 1: $$\int_{0}^{1} 1 dx = \left[ x \right]_0^1 = (1 - 0) = 1$$ So the definite integral is equal to 1: $$\int_{0}^{1} \frac{16^x}{4^{2x}} dx = 1.$$

Key concepts

Integral Evaluation

When we talk about integral evaluation, we're discussing the process of finding the total accumulation of a function over a given interval. This is reflected in the exercise where we were asked to evaluate a definite integral from 0 to 1. A definite integral represents the area under the curve of the function between two points on the x-axis. It's like measuring the total growth or decline across those points.

Ingram evaluation, we begin by simplifying the problem to an easily workable format. Then, we proceed to find the antiderivative of the function, which provides a general function whose rate of change (or derivative) is equal to the original function we're considering. Finally, we apply the fundamental theorem of calculus, which connects derivatives to integrals, to evaluate this at the upper and lower bounds of our interval.

Antiderivatives

An antiderivative of a function is another function that reverses the process of differentiation. If you have the derivative of a function, finding the antiderivative means you're determining the original function before it was differentiated. In the context of integral calculus, antiderivatives play a critical role as they allow us to evaluate definite integrals using the bounds given.

Once the integrand is simplified, we proceed to find the antiderivative. In the example given, after simplifying the integrand to 1, the antiderivative is simply x. This is because the derivative of x with respect to x is 1. Thus, evaluating the antiderivative between the bounds 0 and 1, we identify the area under the function represented by the definite integral.

Exponent Rules

Understanding exponent rules is key when wrestling with integrals involving exponential functions. Exponent rules are the guidelines that explain how to handle expressions with powers. Here are a few critical rules that apply to the exercise:

• Product of Powers: To multiply two powers that have the same base, you add the exponents. For instance, \( a^m \cdot a^n = a^{m+n} \).
• Power of a Power: To raise a power to another power, you multiply the exponents. For example, \( (a^m)^n = a^{mn} \).
• Division of Powers: To divide two powers with the same base, you subtract the exponents. As seen in \( a^m / a^n = a^{m-n} \).

Applying these rules can simplify integrands significantly, as demonstrated in the sample exercise where \(16^x/4^{2x}\) was simplified using these exponent rules.

Simplifying Integrands

The concept of simplifying integrands demands altering the integrand into a more manageable form before tackling the integral. This step is vitally crucial because it can transform a complex-looking integral into something much simpler, as was the case in the provided exercise.

By applying exponent rules, the seemingly complicated expression \(16^x/4^{2x}\) was simplified to 1, a constant. The process of simplification is strategic, aimed at making the evaluation of the integral more direct and less error-prone. While some might overlook this step as merely procedural, it's indeed foundational to efficient problem-solving in calculus. This simplification leads us directly to an easily computed antiderivative, which is the next critical step towards finding the definite integral's value.