Problem 81 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of the integral \(\int_{1}^{2} \sqrt{(2 x+3)\left(3 x^{2}+4\right)} d x\) cannot exceed (A) \(\sqrt{48}\) (B) \(\sqrt{66}\) (C) \(\sqrt{73}\) (D) None of these

Step-by-Step Solution

Answer

The value cannot exceed \(\sqrt{73}\). The correct answer is option (C).

Step 1: Understand the Problem

We need to evaluate the integral \(\int_{1}^{2} \sqrt{(2x+3)(3x^{2}+4)}\, dx\) and find an upper bound for its value. The options provided represent possible maximum values that the integral may reach.

Step 2: Estimate the Expression

To find an upper bound for the integral, we estimate the maximum value of the integrand \(f(x)=\sqrt{(2x+3)(3x^2+4)}\) over the interval from 1 to 2. Calculate the function at the endpoints and consider any increase or decrease within the interval.

Step 3: Calculate the Function at Endpoints

Evaluate the expression at \(x=1\): \(\sqrt{(2\cdot1+3)(3\cdot1^2+4)} = \sqrt{(5)(7)} = \sqrt{35}\). Evaluate at \(x=2\): \(\sqrt{(2\cdot2+3)(3\cdot2^2+4)} = \sqrt{(7)(16)} = \sqrt{112}\).

Step 4: Analyze the Behavior of the Function

Since the function is increasing from \(x=1\) to \(x=2\), \(\sqrt{112}\) is the maximum value over this interval. Thus, the upper bound of the integral will be close to the maximum value as it stretches from 1 to 2.

Step 5: Evaluate the Integral's Upper Bound

The integral's value is less than or equal to the integral of the maximum function value over the interval. Calculate the area: \(\text{Max value}\times\text{Length of interval} = \sqrt{112} \times (2-1) = \sqrt{112} = \sqrt{16\times 7} = 4\sqrt{7}\approx 10.583\). This is less than \(\sqrt{66}\approx 8.124\), so option (B) \(\sqrt{66}\) is incorrect. Checking option (C) is unnecessary since it's larger.

Key Concepts

Definite Integrals

In calculus, a definite integral is a way to calculate the accumulation of quantities, like area under a curve, over an interval. It is expressed in the form \[\int_{a}^{b} f(x) \, dx,\]where \(a\) and \(b\) are the limits of integration and \(f(x)\) is the function being integrated. Definite integrals help us find the total accumulation from \(a\) to \(b\).

• \(\int_{1}^{2} \sqrt{(2x+3)(3x^{2}+4)} \, dx\) is a definite integral with limits 1 and 2.
• The function \(f(x) = \sqrt{(2x+3)(3x^{2}+4)}\) is the integrand, which is the target of our integration.
• Definite integrals differ from indefinite ones, as they result in a numerical value instead of a function with a 'C'.

In practical terms, definite integrals often represent quantities such as distances or areas based on the context of the problem.

Square Roots in Integrals

When solving integrals that contain square roots, like \[f(x) = \sqrt{(2x+3)(3x^2+4)},\]it’s important to correctly handle the complexity they introduce. Square roots often require techniques beyond simple integration rules.

• Simplifying expressions inside the square root can sometimes help, but this can be complex depending on the problem.
• Accurate computation of values at key points, like the boundaries of integration, is crucial for understanding the integral's behavior.

In the given exercise, evaluating \(f(x)\) at \(x=1\) and \(x=2\) provided essential information about the function's maximum value on the interval [1,2]. Understanding how square roots affect an integrand's complexity helps in determining limits and behavior of functions.

Integral Estimation

Integral estimation finds approximations rather than calculating exact values, particularly when exact solutions are challenging. This involves assessing the behavior of the integrand:

• Approximate the behavior of \(f(x)\) over the interval, which in this case is [1,2].
• Check function values at endpoints to gauge its increase or decrease.
• Decide on potential upper bounds based on known function values.

In the exercise, understanding that \(f(x)\) increases and reaches a maximum at \(x=2\), the maximum value \(\sqrt{112}\) was used to establish the possible upper limit of our integral. Then multiplying this by the interval length estimates the area under the curve without exact integration. Comparing this with provided choices helps decide the best approximation.