Evaluate the integrals in Exercises 87–96.
89. ∫₀¹ 2^(−θ) dθ

Verified step by step guidance

1. Identify the integral to be evaluated: \(\int_0^1 2^{-\theta} \, d\theta\).
2. Rewrite the integrand using the exponential function with base \(e\): \(2^{-\theta} = e^{-\theta \ln(2)}\).
3. Set up the integral with the exponential form: \(\int_0^1 e^{-\theta \ln(2)} \, d\theta\).
4. Integrate using the formula for the integral of an exponential function: \(\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C\). Here, \(a = -\ln(2)\).
5. Evaluate the definite integral by substituting the limits \(\theta = 0\) and \(\theta = 1\) into the antiderivative and subtracting: \(\left[ \frac{e^{-\theta \ln(2)}}{-\ln(2)} \right]_0^1\).

Key Concepts

Definite Integral

A definite integral calculates the net area under a curve between two limits. It is represented as ∫_a^b f(x) dx, where a and b are the bounds. Evaluating a definite integral involves finding the antiderivative and then applying the Fundamental Theorem of Calculus.

Exponential Functions with Variable Exponents

Exponential functions like 2^(-θ) have a constant base and a variable exponent. Understanding how to integrate such functions requires rewriting them using natural logarithms or recognizing their derivative patterns, as the integral of a^x involves dividing by ln(a).

Fundamental Theorem of Calculus

This theorem links differentiation and integration, stating that if F is an antiderivative of f, then ∫_a^b f(x) dx = F(b) - F(a). It allows evaluation of definite integrals by computing the difference of antiderivative values at the bounds.