91Ó°ÊÓ

Definite integrals are a fundamental concept in calculus, allowing us to calculate the net area under a curve between two specific points, known as the bounds. The given problem requires evaluating\[ \int_{\pi^{2}}^{4\pi^{2}} \frac{1}{\sqrt{x}} \sin \sqrt{x} \; dx \]which involves computing the area from \(x = \pi^2\) to \(x = 4\pi^2\). Definite integrals have both an upper and lower limit which are used to find the overall result of the integral. The process to evaluate definite integrals involves:

• Finding the antiderivative of the function within the integral.
• Substituting the upper and lower bounds into the antiderivative.
• Subtracting to find the result.

In this problem, solving the integral with the bounds provided simplifies to applying these steps, resulting in the integral evaluation giving a value of zero, meaning there is no net area within these bounds.

The substitution method, often called "u-substitution," serves as a vital tool for simplifying complex integrals. It is essentially the reverse of the chain rule for derivatives. In this problem, the integral \[ \int_{\pi^{2}}^{4\pi^{2}} \frac{1}{\sqrt{x}} \sin \sqrt{x} \, dx \]is made more manageable by setting \( u = \sqrt{x} \). This substitution transforms the integrand from a complex form into a simpler one:

• \( x = u^2 \), therefore \( dx = 2u \, du \).
• The integral's limits change: when \( x = \pi^2 \), \( u = \pi \), and when \( x = 4\pi^2 \), \( u = 2\pi \).
• Thus, the original integral becomes \( 2 \int_{\pi}^{2\pi} \sin u \, du \).

The substitution effectively reduces the integral to a basic trigonometric integral, which can be easily evaluated. This method is particularly useful when dealing with integrals involving square roots or composite functions, as it turns them into something more straightforward to solve.

Trigonometric integrals involve integrals with sine, cosine, and other trigonometric functions. In our exercise, once we perform u-substitution, the integral with trigonometric elements becomes much clearer:\[ \int \sin u \, du \]This type of integral can be easily solved as the antiderivative of \( \sin u \) is \( -\cos u \). Understanding trigonometric integrals involves recognizing patterns and antiderivatives, which simplifies the computing process significantly. Once the antiderivative is found, the definite integral evaluation continues:

• Compute \(-\cos u\) at both limits \( \pi \) and \( 2\pi \).
• Use \( \cos(2\pi) = 1 \) and \( \cos(\pi) = -1 \) in the evaluation.
• The calculation simplifies to using the identities and subtracting appropriately to find the final value.

Trigonometric integrals are a common occurrence in calculus, and mastering them bolsters overall competence in solving diverse integral types.