91Ó°ÊÓ

Definite integrals allow us to calculate the area under a curve within a specific interval. In calculus, these integrals are defined by two limits, often denoting the start and end points over which we integrate a function.
For example, consider the definite integral \( \int_{0}^{x^2} \sec^2 t \, dt \). This expression represents the area under the curve of \( \sec^2 t \) from \( t = 0 \) to \( t = x^2 \).
The process involves finding an antiderivative of the function inside the integral. In our exercise, the antiderivative of \( \sec^2 t \) is \( \tan t \). Therefore,

• Compute the antiderivative: \( \tan t \)
• Evaluate from 0 to \( x^2 \): \( \tan(x^2) - \tan(0) = \tan(x^2) \) since \( \tan(0) = 0 \)

This method of using definite integrals helps in many fields, such as physics, where they are used to determine work done by a force over a distance.

Trigonometric functions, such as sine, cosine, and tangent, play a crucial role in calculus, especially in handling limits and integrals.
When dealing with expressions like \( \tan(x^2) \) or \( x \sin x \), it's essential to understand their behavior as \( x \rightarrow 0 \). In calculus:

• \( \tan(x^2) \) approaches \( x^2 \) because for small angles, tangent can be approximated as the angle itself in radians.
• Similarly, \( \sin x \) can be approximated by \( x \) for small values, due to the limit property \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).

These approximations simplify complicated trigonometric expressions, making them easier to evaluate, particularly in limit scenarios.

L'Hôpital's Rule is a powerful tool in calculus used to evaluate limits of indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When faced with such forms, we differentiate the numerator and the denominator separately.
In the exercise, after simplifying the expression to \( \frac{\tan(x^2)}{x \sin x} \), both the numerator and the denominator approach zero as \( x \rightarrow 0 \). Hence, it's a perfect scenario to apply L'Hôpital's Rule.

• Differentiate the numerator: \( (\tan(x^2))' = \sec^2(x^2) \cdot 2x \).
• Differentiate the denominator: \( (x \sin x)' = \sin x + x \cos x \).

After implementing L'Hôpital's Rule, evaluate the limit again. Simplifying \( \frac{2x \sec^2(x^2)}{\sin x + x \cos x} \), we find it approaches 1 as \( x \rightarrow 0 \). Understanding L'Hôpital's Rule allows us to solve complex limits efficiently, validating their evaluations with precise calculus techniques.