Class CBSE Class 12 Mathematics Definite Integrals Q #629
KNOWLEDGE BASED
APPLY
1 Marks 2024 AISSCE(Board Exam) MCQ SINGLE
The value of \(\int_{-1}^{1}x|x|dx\) is:
(A) \(\frac{1}{6}\)
(B) \(\frac{1}{3}\)
(C) \(-\frac{1}{6}\)
(D) 0
Prev Next

AI Tutor Explanation

Powered by Gemini

Step-by-Step Solution

The given integral is \(\int_{-1}^{1}x|x|dx\).

We can split the integral into two parts based on the definition of \(|x|\):

\(|x| = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } x \geq 0 \end{cases}\)

So, the integral becomes:

\(\int_{-1}^{1}x|x|dx = \int_{-1}^{0}x(-x)dx + \int_{0}^{1}x(x)dx\)

\(= \int_{-1}^{0}-x^2 dx + \int_{0}^{1}x^2 dx\)

\(= \left[-\frac{x^3}{3}\right]_{-1}^{0} + \left[\frac{x^3}{3}\right]_{0}^{1}\)

\(= \left(-\frac{0^3}{3} - \left(-\frac{(-1)^3}{3}\right)\right) + \left(\frac{1^3}{3} - \frac{0^3}{3}\right)\)

\(= \left(0 - \frac{1}{3}\right) + \left(\frac{1}{3} - 0\right)\)

\(= -\frac{1}{3} + \frac{1}{3}\)

\(= 0\)

Correct Answer: 0

AI Suggestion: Option D

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the definition of absolute value and the rules of integration to solve the problem.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure to evaluate the definite integral, including splitting the integral based on the absolute value function and then applying the power rule of integration.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of integration techniques and properties of definite integrals as covered in the textbook.