Problem 87 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(83-92.\) $$\int_{1}^{\sqrt{2}} x 2^{\left(x^{2}\right)} d x$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{\ln 2} \).

Step 1: Identify the Integral Type

The integral given is \( \int_{1}^{\sqrt{2}} x 2^{(x^2)} \, dx \). This appears to be a non-elementary integral due to the exponential containing \( x^2 \). Therefore, standard techniques like substitution or integration by parts might be tried, but eventually, it might need special functions or numerical methods.

Step 2: Attempt Substitution

Try a substitution that simplifies the exponent. Let \( u = x^2 \). Then \( du = 2x \, dx \) or \( x \, dx = \frac{1}{2} \, du \). Rewrite the integral in terms of \( u \).Change the limits of integration: when \( x = 1 \), \( u = 1^2 = 1 \); when \( x = \sqrt{2} \), \( u = (\sqrt{2})^2 = 2 \). Thus, the integral becomes \( \frac{1}{2} \int_{1}^{2} 2^u \, du \).

Step 3: Integrate with Respect to u

The integral \( \int 2^u \, du \) can be integrated using the formula for the integral of an exponential function: \[ \int a^u \, du = \frac{a^u}{\ln a} + C, \] where \( a \) is a constant. In this case, \( a = 2 \), so:\[ \int 2^u \, du = \frac{2^u}{\ln 2}. \]

Step 4: Evaluate the Integral

Now, place the antiderivative back into the integral with the boundaries:\[ \frac{1}{2} \left[ \frac{2^u}{\ln 2} \right]_{1}^{2} \].This can be calculated as:\[ \frac{1}{2} \left[ \frac{2^2}{\ln 2} - \frac{2^1}{\ln 2} \right] = \frac{1}{2} \left[ \frac{4}{\ln 2} - \frac{2}{\ln 2} \right] = \frac{1}{2} \left[ \frac{4 - 2}{\ln 2} \right] = \frac{1}{2} \cdot \frac{2}{\ln 2} = \frac{1}{\ln 2}. \]

Step 5: Calculate the Final Answer

The value of \( \frac{1}{\ln 2} \) is a constant, approximately 1.4427. Thus, the final evaluated integral is \( \frac{1}{\ln 2} \).

Key Concepts

Non-elementary Integrals

In integral calculus, non-elementary integrals represent those integrals that cannot be expressed in terms of elementary functions. Elementary functions include polynomial, rational, trigonometric, exponential, and logarithmic functions along with their inverses. These integrals often involve complex expressions involving exponential, logarithmic, or trigonometric forms in non-standard arrangements that traditional methods cannot easily simplify.

For instance, the integral \( \int x 2^{(x^2)} \, dx \) is non-elementary because the exponential function combines with a squared variable. This complexity arises because there is no straightforward antiderivative that falls under the "elementary" category for this expression.

To evaluate non-elementary integrals, mathematicians might apply methods such as substitution or integration by parts. Alternatively, some integrals might require numerical methods or special functions, like the Gamma function or Error function, depending on their complexity. Recognizing when an integral is non-elementary efficiently guides mathematicians towards choosing the best method or tool for the job.

Substitution Method

The substitution method is a powerful technique in integral calculus for making complex integrals manageable. This technique involves changing the variable of integration to simplify the integral's formula. The essence is to recognize part of the integrand that can be replaced with a new variable.

In our example, \( \int x 2^{(x^2)} \), substitution becomes an effective approach. By letting \( u = x^2 \), the integral transforms into an expression in terms of \( u \). This change is based on the derivative \( du = 2x \, dx \), simplifying to \( x \, dx = \frac{1}{2} \, du \).

• Redefine the integral limits. When \( x = 1 \), \( u = 1^2 = 1 \). As \( x = \sqrt{2} \), \( u = 2 \).
• The integral becomes simpler: \( \frac{1}{2} \int_{1}^{2} 2^u \, du \).

This method helps not only in simplifying the calculations but also in yielding more elegant forms which can be approached using standard integral formulas.

Exponential Functions

Exponential functions are fundamental in calculus, represented generally as \( a^x \), where \( a \) is a positive constant. These functions illustrate rapidly growing behavior as the power of the exponent increases, making them crucial in modeling exponential growth and decay in various fields.

To integrate exponential functions like \( \int 2^u \, du \), one applies a specific formula:\[\int a^u \, du = \frac{a^u}{\ln a} + C\]Here, \( a \) represents the base of the exponential function. In our scenario, with \( a = 2 \), the integration becomes:\[\int 2^u \, du = \frac{2^u}{\ln 2}\]This formula accounts for the fact that the natural logarithm of \( a \) scales the integral to accommodate the growth rate of the exponential. Additionally, integration constants \( C \) are conventionally included for indefinite integrals, but omitted for definite integrals as we proceed to evaluate the boundary differences.