Evaluate the integrals in Exercises 31–78.
58. ∫(from 0 to ln9)e^θ(e^θ-1)^(1/2) dθ

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Start by examining the integral: \(\int_0^{\ln 9} e^{\theta} \left(e^{\theta} - 1\right)^{1/2} \, d\theta\). Notice that the integrand contains \(e^{\theta}\) and a function of \(e^{\theta} - 1\), which suggests a substitution involving \(e^{\theta}\).

Let’s perform the substitution \(u = e^{\theta} - 1\). Then, differentiate both sides with respect to \(\theta\) to find \(du\): \(du = e^{\theta} d\theta\).

Rewrite the integral in terms of \(u\). Since \(du = e^{\theta} d\theta\), the integral becomes \(\int \sqrt{u} \, du\) after changing the limits accordingly.

Change the limits of integration from \(\theta\) to \(u\): when \(\theta = 0\), \(u = e^0 - 1 = 0\); when \(\theta = \ln 9\), \(u = e^{\ln 9} - 1 = 9 - 1 = 8\).

Now, the integral is \(\int_0^8 u^{1/2} \, du\). This is a standard power integral, which can be integrated using the formula \(\int u^n du = \frac{u^{n+1}}{n+1} + C\) for \(n \neq -1\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integrals

A definite integral calculates the net area under a curve between two specified limits. It involves evaluating the antiderivative at the upper and lower bounds and subtracting these values. This process yields a numerical value representing the accumulated quantity over the interval.

Substitution Method

The substitution method simplifies integrals by changing variables to transform the integral into a more manageable form. By letting a new variable equal a function inside the integral, the integral often becomes easier to evaluate. This technique is especially useful when the integral contains composite functions.

Exponential Functions and Their Properties

Exponential functions, like e^θ, have unique properties such as their derivative and integral being proportional to themselves. Understanding how to manipulate expressions involving exponentials, including their growth and composition with other functions, is essential for solving integrals involving these terms.

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