KNOWLEDGE BASED
APPLY
1 Marks
2025
AISSCE(Board Exam)
MCQ SINGLE
The area of the shaded region (figure) represented by the curves \(y=x^{2}\), \(0\le x\le2\) and y-axis is given by
(A)
\(\int_{0}^{2} x^2 dx\)
(B)
\(\int_{0}^{2} \sqrt{y} dy\)
(C)
\(\int_{0}^{4} x^2 dx\)
(D)
\(\int_{0}^{4} \sqrt{y} dy\)
Correct Answer: D
Explanation
\(\int_{0}^{4} \sqrt{y} dy\)
Step-by-Step Solution
To find the area of the shaded region, we need to integrate the function with respect to the appropriate variable and limits. The region is bounded by the curve \(y = x^2\), the y-axis, and the lines \(x = 0\) and \(x = 2\). Since the region is defined with respect to the y-axis, we need to express x in terms of y. From \(y = x^2\), we have \(x = \sqrt{y}\). The limits of integration for y will be from \(y = 0\) (when \(x = 0\)) to \(y = 4\) (when \(x = 2\)). Therefore, the area of the shaded region is given by the integral: \[ \int_{0}^{4} \sqrt{y} \, dy \]Correct Answer: \(\int_{0}^{4} \sqrt{y} dy\)<\/strong>
Pedagogical Audit
Bloom's Analysis:
This is an APPLY question because it requires the student to apply the concept of definite integrals to calculate the area under a curve. The student needs to identify the correct limits of integration and the correct function to integrate.
Knowledge Dimension:
PROCEDURAL
Justification:
The question requires the student to apply a specific procedure (integration) to solve the problem. It involves knowing how to set up the integral for the area under a curve and how to evaluate it.<\/span>
Syllabus Audit:
In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding and application of a standard textbook method for finding the area under a curve using integration.
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