Problem 71 - AP Calculus with 8 Practice Tests 15th

Question

\(\int e^{2 \ln u} d u=\) (A) \(\frac{1}{3} e^{u^{3}}+C\) (B) \(e^{u^{3} / 3}+C\) (C) \(\frac{1}{3} u^{3}+C\) (D) \(\frac{2}{u} e^{2 \ln u}+C\)

Step-by-Step Solution

Answer

The answer is (C) \( \frac{1}{3} u^3 + C \).

Step 1: Simplify the Exponent

The given integral is \( \int e^{2 \ln u} \, du \). Using the property of logarithms, we know \( e^{2 \ln u} = (e^{\ln u})^2 = u^2 \). So the integral simplifies to \( \int u^2 \, du \).

Step 2: Integrate the Monomial

Now that we have the simplified integral \( \int u^2 \, du \), we use the power rule for integration. The power rule states that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), where \( n eq -1 \). For \( n = 2 \), we get \( \int u^2 \, du = \frac{u^{2+1}}{2+1} + C = \frac{u^3}{3} + C \).

Key Concepts

Power Rule for Integration

The power rule for integration is a fundamental technique in integral calculus. It is among the first rules to master for solving integrals involving powers of a variable. This rule is straightforward:

• If you are integrating a function of the form \( u^n \), where \( u \) is the variable and \( n \) is any real number except -1, you apply the power rule.
• It states that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.
• This rule allows you to easily find the antiderivative of polynomial expressions.

In the previous exercise, once we simplified the expression to \( u^2 \), the power rule was applied with \( n=2 \). This resulted in the integrated expression \( \frac{u^3}{3} + C \).Understanding and applying the power rule is essential for handling basic and more complex integrals that appear in calculus.

Logarithmic Properties

The properties of logarithms are vital tools in simplifying expressions within integrals. Logarithms can transform products into sums, powers into multiples, and can even help in reversing an exponentiation resulting in a simplification.Consider the expression \( e^{2 \ln u} \). With the property of logs and exponentials, we use:

• \( e^{\ln a} = a \), since the exponential and the natural logarithm are inverse functions.
• Therefore, \( e^{2 \ln u} = (e^{\ln u})^2 = (u)^2 = u^2 \).

By simplifying the expression this way, integrating the function becomes much more straightforward. Recognizing and applying these logarithmic properties can reduce complexity and make integral problems less daunting.

Exponent Simplification

Simplifying exponents is another crucial skill in calculus. Not only does it make expressions easier to handle, but it also prepares them for integration or differentiation.In the given exercise, we started with \( e^{2 \ln u} \). This expression seemed complex at first glance, but simplifying it using exponent rules made our work easier.

• We utilized the property \( e^{\ln u} = u \), and hence \( e^{2 \ln u} = \left( e^{\ln u} \right)^2 = u^2 \).
• This simplification transformed the original integral into a more approachable \( \int u^2 \, du \).

Exponent simplification leverages laws like power rules and properties that relate logarithms with exponentials. This simplification is critical for correctly solving calculus problems and optimizing the integration process.