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

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Identify the integral to be evaluated: \(\int_0^1 2^{-\theta} \, d\theta\).

Rewrite the integrand using the exponential function with base \(e\): \(2^{-\theta} = e^{-\theta \ln(2)}\).

Set up the integral with the exponential form: \(\int_0^1 e^{-\theta \ln(2)} \, d\theta\).

Integrate using the formula for the integral of an exponential function: \(\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C\). Here, \(a = -\ln(2)\).

Evaluate the definite integral by substituting the limits \(\theta = 0\) and \(\theta = 1\) into the antiderivative and subtracting: \(\left[ \frac{e^{-\theta \ln(2)}}{-\ln(2)} \right]_0^1\).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.

Evaluate the integrals in Exercises 87–96.89. ∫₀¹ 2^(−θ) dθ