Problem 105 - Applied Calculus

Question

Evaluate \(\int_{1}^{1} \frac{x^{43} e^{-17 x}+219 \sqrt[3]{x^{2}}}{\ln \sqrt[29]{6 x^{3}-x^{-11}}-\pi^{3}} d x .\) [Hint: No work necessary.

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Identify the Integral Limits

First, note the limits of integration: both the upper and lower limits of the integral are the same (1 to 1).

Step 2: Apply the Property of Definite Integrals

According to the property of definite integrals, if the upper and lower limits are the same, the value of the integral is 0, regardless of the function being integrated.

Key Concepts

Properties of Integrals

Understanding the properties of integrals is a fundamental part of mastering calculus. One key property is that if the upper and lower limits of integration are the same, the result of the integral is zero. This property holds regardless of how complex the function may look within the integral.
In the context of the exercise, the integrand \( \frac{x^{43} e^{-17 x}+219 \sqrt[3]{x^{2}}}{\ln \sqrt[29]{6 x^{3}-x^{-11}}-\pi^{3}} \) seems daunting. However, due to the same-limits property, the focus is on the integration limits rather than the integrand itself.

• If the limits a and b are equal, then \( \int_{a}^{a} f(x) \, dx = 0 \). This is true despite the function \( f(x) \).

The fact that integration with identical upper and lower bounds renders any function’s area under the curve nonexistent simplifies many problems. Recognizing and applying properties like these can help streamline solving complex calculus problems.

Integral Limits

Integral limits define the bounds within which we evaluate the area under a curve represented by the function. In definite integrals, these bounds are critical as they directly influence the result of the computation.

Generally, the definite integral \( \int_{a}^{b} f(x) \, dx \) calculates the net area between the curve \( f(x) \)and the x-axis from \( x = a \)to \( x = b \).

• If the limits are the same, by property, the integral equals zero.
• Understanding limits allows you to determine the domain of integration and focus on the relevant section of the function.

The integral limits not only dictate the portion of the function being considered but also simplify calculations significantly when they hold symmetries or particular values, as seen in the exercise example. Identifying these specific cases is very helpful in calculus problem-solving.

Calculus Problem-Solving

Successfully solving calculus problems involves both recognizing patterns and applying theorems effectively. When approaching integrations, it’s vital to remember that the nature of calculus allows us to utilize specific properties to simplify unnecessarily complex tasks.

In this exercise, understanding the properties of integrals led us directly to the solution without unnecessary work. Here are some strategies in calculus problem-solving:

• Always examine the limits of the integral first as they can simplify the task immensely, like leading to a zero outcome when they match.
• Identify and apply fundamental properties and theorems which can streamline your work.
• Analyze the integrand to determine if substitution or any specific theorem could apply aside from basic property use.

By breaking down the problem, focusing on essentials like limits, and applying useful properties, calculus problem-solving can become a methodical, less daunting task. These skills develop with practice and a clear understanding of fundamental calculus concepts.