Problem 180 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\int_{0}^{2}\left[x^{2}\right] d x\) is \(\quad[2002]\) (A) \(2-\sqrt{2}\) (B) \(2+\sqrt{2}\) (C) \(\sqrt{2}-1\) (D) \(-\sqrt{2}-\sqrt{3}+5\)

Step-by-Step Solution

Answer

The provided options do not match the correct result, \( \frac{8}{3} \). Check the problem again.

Step 1: Identify the function to be integrated

The given function to integrate is \( f(x) = x^2 \), and we are tasked with finding the definite integral of \( x^2 \) from 0 to 2.

Step 2: Set up the definite integral

Set up the integral: \[ \int_{0}^{2} x^2 \, dx \].

Step 3: Find the antiderivative

The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \).

Step 4: Evaluate the antiderivative at the bounds

First, evaluate at the upper bound 2: \[ \frac{2^3}{3} = \frac{8}{3} \]. Next, evaluate at the lower bound 0: \[ \frac{0^3}{3} = 0 \].

Step 5: Calculate the definite integral

Subtract the evaluation at the lower bound from the upper bound: \[ \frac{8}{3} - 0 = \frac{8}{3} \].

Step 6: Compare result with given options

Out of the options provided, none of them match the result \( \frac{8}{3} \). Ensure the problem and options are correctly interpreted, as there might be an error in the options provided.

Key Concepts

Antiderivative

When dealing with integrals, the antiderivative is a powerful tool. It is essentially the reverse process of differentiation. Consider it as finding a function whose derivative is the given function. For the function \(f(x) = x^2\), we need to find a function \(F(x)\) such that \(F'(x) = x^2\).
To find the antiderivative of \(x^2\), we use the basic power rule for integrals, which states: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration.
Applying this rule, the antiderivative of \(x^2\) becomes \(\frac{x^3}{3} + C\). In the context of definite integrals, the constant \(C\) is not needed as it cancels out during the integration process.

Definite Integral Calculation

Definite integrals allow us to compute the area under a curve within specified bounds. Once the antiderivative is found, calculating the definite integral over a range of values becomes straightforward. Consider the integral \(\int_0^2 x^2 \, dx\).
First, identify the antiderivative, which from our previous section, is \(\frac{x^3}{3}\).
Then, apply the Second Fundamental Theorem of Calculus, which states that to find the definite integral, you need to evaluate the antiderivative at the upper and lower bounds of the integral, and subtract the two results:

• Evaluate \(\frac{x^3}{3}\) at the upper bound 2: \(\frac{2^3}{3} = \frac{8}{3}\).
• Evaluate at the lower bound 0: \(\frac{0^3}{3} = 0\).
• Subtract the evaluations: \(\frac{8}{3} - 0 = \frac{8}{3}\).

This gives us the exact area under \(x^2\) from 0 to 2.

Bounds Evaluation

Bounds evaluation is a crucial step in determining the value of a definite integral. Carefully handling the upper and lower limits ensures the process is accurate from start to finish.
In our exercise, the bounds were 0 and 2. These values act as limits to evaluate the antiderivative. Starting with the upper bound, you substitute 2 into the antiderivative function \(\frac{x^3}{3}\) and perform the calculation to find the exact output.
Next, perform the same substitution for the lower bound value, 0. As \(\frac{0^3}{3} = 0\), this simplifies the process since any number raised to a power of 0 and divided remains zero.
Finally, the result is found by subtracting the lower bound evaluation from the upper bound's value. Proper handling of these bounds is vital to the accuracy of the integrated value. It helps us ensure that no steps are overlooked in the calculation process.