Problem 105 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\int_{0}^{1} \frac{d x}{1+x^{2}+2 x^{5}}\) lies between (A) \(\frac{\pi}{6 \sqrt{3}}\) and \(\frac{\pi}{4}\) (B) \(\frac{\pi}{3 \sqrt{3}}\) and \(\frac{\pi}{2}\) (C) \(\frac{\pi}{3 \sqrt{3}}\) and \(\frac{\pi}{4}\) (D) None of these

Step-by-Step Solution

Answer

The integral lies between \( \frac{\pi}{6 \sqrt{3}} \) and \( \frac{\pi}{4} \), so option (A) is correct.

Step 1: Recognize the Integral Form

The given integral is \( \int_{0}^{1} \frac{d x}{1+x^{2}+2 x^{5}} \). We need to estimate the value of this integral to determine which interval it falls into.

Step 2: Consider the Simplified Version

To estimate, interpret the function \( f(x) = \frac{1}{1+x^2+2x^5} \). Notice that for \( x \geq 0 \), \( 2x^5 \geq 0 \), hence \( f(x) \leq \frac{1}{1+x^2} \). Logical bounds will help ascertain the placement without precise integration.

Step 3: Simplified Integration for Upper Bound

Since \( f(x) \leq \frac{1}{1+x^2} \), take \( \int_{0}^{1} \frac{d x}{1+x^2} \). This integral evaluates to \( \left[ \tan^{-1}(x) \right]_{0}^{1} = \frac{\pi}{4} \). The upper bound of the integral is thus \( \frac{\pi}{4} \).

Step 4: Consider Lower Bound Estimation

Now observe if \( f(x) \geq \frac{1}{1+x^2+2} \). Thus consider \( \int_{0}^{1} \frac{d x}{1+x^2+2} = \int_{0}^{1} \frac{d x}{3+x^2} \). This is of form \( \int \frac{d x}{a+x^2} \) which evaluates to \( \frac{1}{\sqrt{a}} \tan^{-1}(\frac{x}{\sqrt{a}}) \) for \( a > 0 \).

Step 5: Evaluate the Lower Bound Integral

Given \( a = 3 \), it follows \( \frac{1}{\sqrt{3}} \left[ \tan^{-1}(\frac{x}{\sqrt{3}}) \right]_{0}^{1} = \frac{1}{\sqrt{3}} \times (\tan^{-1}(\frac{1}{\sqrt{3}})) = \frac{1}{\sqrt{3}} \times \frac{\pi}{6} = \frac{\pi}{6 \sqrt{3}} \). The lower bound of the integral is thus \( \frac{\pi}{6 \sqrt{3}} \).

Step 6: Conclusion

The integral \( \int_{0}^{1} \frac{d x}{1+x^{2}+2 x^{5}} \) lies between \( \frac{\pi}{6 \sqrt{3}} \) and \( \frac{\pi}{4} \). This falls under option (A).

Key Concepts

Integral Bounds

When working with definite integrals, understanding the concept of bounds can provide you with a great estimation of where the value of the integral lies without computing it exactly. In this exercise, the bounds are essential because they give us an interval within which the actual integral value exists. Here's how that works:

• For function estimation, use inequalities to create a simpler function that can be readily integrated.
• Compare this simplified function to the original to establish upper and lower bounds.

For our integral, we have evaluated it between two bounds to ascertain its approximate value range. This approach is handy when exact integration is complex or time-consuming.

Trigonometric Substitution

In calculus, a fantastic technique to simplify the process of integration is trigonometric substitution. Though in this particular problem we did not deeply explore trigonometric substitution, it is worth noting how valuable this method can be. Typically, this technique is used to

• Convert a complex function involving square roots into a simpler trigonometric function.
• Simplify the process of integration by transforming the variable of integration to a trigonometric variable, often making the integral more manageable.

While this exercise did not require trigonometric substitution explicitly, the studied form of \( \tan^{-1}(x) \)—a result linked to trigonometric considerations—illustrates the pervasive influence of trigonometric concepts in integration solutions.

Integral Estimation

Integral estimation is a practical and efficient way of obtaining approximate values of an integral when precise computation isn't feasible. Estimation involves:

• Recognizing simpler forms of the given function, which can act as envelopes for the actual function over a certain interval.
• Acknowledging that these forms offer bounds within which the integral value will lie, without calculating it exactly.

In this problem, we utilized estimation by comparing the function to simpler integrals that could be directly evaluated. Through this method, the integral was estimated to have a value between \( \frac{\pi}{6 \sqrt{3}} \) and \( \frac{\pi}{4} \), demonstrating how efficient integral estimation can be for tackling challenging calculus problems.

Function Inequality

Understanding inequalities of functions is crucial when defining the behavior of an integral. Function inequalities provide:

• The means to bound the original function by simpler functions above and below, reflecting potential minimum and maximum values over the integration interval.
• A practical approach to comparing and determining which functions are greater or lesser than the original function over a certain domain.

In this exercise, we used the inequality \( f(x) \leq \frac{1}{1+x^2} \) and \( f(x) \geq \frac{1}{1+x^2+2} \) to draw conclusions about the integral. These inequalities allowed us to define where the value of the integral lies, without the need for calculating the exact integral itself. This technique is powerful for simplifying and solving integrals where direct methods prove cumbersome.