Problem 104 - A Complete Resource Book in Mathematics for JEE Main

Question

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

Step-by-Step Solution

Answer

The integral lies between \(\frac{1}{4}\) and 1, so the answer is (A).

Step 1: Understand the Function

The given function for integration is \( f(x) = \frac{1}{1+x^2+2x^5} \). We need to determine the bounds for the integral \( \int_{0}^{1} f(x) \, dx \).

Step 2: Determine Lower and Upper Bounds

Since \( x \) is in the range from 0 to 1, the expression \( 1 + x^2 + 2x^5 \) varies between 1 and 4. Therefore, \( \frac{1}{4} \) and 1 act as the bounds for \( \frac{1}{1 + x^2 + 2x^5} \).

Step 3: Evaluate the Integral Bounds

Compute the integral bounds using inequalities: \[ \int_0^1 \frac{1}{4} \, dx \leq \int_0^1 \frac{1}{1+x^2+2x^5} \, dx \leq \int_0^1 1 \, dx. \]The calculated integrals are \( \frac{1}{4} \) and 1, respectively.

Step 4: Conclusion of Bounds

Since the integral \( \int_0^1 \frac{1}{1+x^2+2x^5} \, dx \) lies between \( \frac{1}{4} \) and 1, we check which option matches these bounds. The correct answer is (A), which is \( \frac{1}{4} \) and 1.

Key Concepts

Definite Integers

In integral calculus, a **definite integral** is an integral with specified upper and lower limits. It calculates the area under the curve of a given function from the lower limit to the upper limit on the x-axis. Helpful when you need to evaluate the total accumulation of quantities, definite integrals are represented by the symbol \( \int_a^b \) where \( a \) and \( b \) are limits. For the exercise, we calculate the definite integral \( \int_0^1 \frac{1}{1+x^2+2x^5} \, dx \). This integral tells us the area under the curve of the function between \( x = 0 \) and \( x = 1 \).

Inequalities

Inequalities are mathematical expressions that show the relationship between different values. Rather than stating that two values are equal, inequalities indicate one value is greater than or less than another.In the exercise solution, inequalities help determine the range of function values within the definite integral. By observing that the expression \( 1 + x^2 + 2x^5 \) ranges between 1 and 4 as \( x \) varies from 0 to 1, we see:

• Minimum value: 1
• Maximum value: 4

Therefore, the function \( \frac{1}{1+x^2+2x^5} \), which is subject to inequalities, points towards the bounds \( \frac{1}{4} \) and 1.

Integral Bounds

Integral bounds refer to the limits set on a definite integral and the boundary in which the function operates. These bounds help in evaluating the integral by providing a clear start and endpoint for the calculation.In our exercise, the integral bounds are from 0 to 1, as in \( \int_0^1 \). These are our x-values for which we analyze the function \( f(x) = \frac{1}{1+x^2+2x^5} \).Using these bounds, inequalities, and properties of the function, we determine that the evaluated integral holds between the limits \( \frac{1}{4} \) and 1.

Function Analysis

Function analysis in calculus involves understanding how a function behaves over given intervals or domains. It often requires examining the function's components, such as its domain, range, and any critical or boundary points.For the function \( \frac{1}{1+x^2+2x^5} \), the major task is to assess its behavior for \(x\) in \([0, 1]\). Since neither the numerator nor the denominator involve any undefined expressions within this interval, we focus purely on the effects of \( x \).Here's what we know:

• The denominator \( 1+x^2+2x^5 \) ranges from 1 to 4 in the tested interval.
• Thus, the fraction's bounds are established as \( \frac{1}{4} \) and 1.

Analyzing both inequalities and integral bounds, we gain insights into how the function's value shifts across its range.