Q1437: Sketch a graph of $y=x^{2}$. Using integration, find the are...

Class CBSE Class 12 Mathematics Applications of Integrals Q #1437

COMPETENCY BASED

UNDERSTAND

5 Marks 2025 AISSCE(Board Exam) LA

Sketch a graph of $y=x^{2}$. Using integration, find the area of the region bounded by $y=9$, $x=0$ and $y=x^{2}$.

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AI Tutor Explanation

Detailed Solution

Step 1: Sketch the graph of $y = x^2$

The graph of $y = x^2$ is a parabola opening upwards with its vertex at the origin (0,0).

Step 2: Identify the region bounded by $y=9$, $x=0$, and $y=x^2$

The region is bounded by the parabola $y=x^2$, the horizontal line $y=9$, and the vertical line $x=0$. We need to find the point of intersection between $y=x^2$ and $y=9$.

Step 3: Find the intersection point of $y=x^2$ and $y=9$

Set $x^2 = 9$. This gives $x = \pm 3$. Since we are bounded by $x=0$, we consider the positive value $x=3$. So the intersection point is (3,9).

Step 4: Set up the integral to find the area

We want to find the area between $y=9$ and $y=x^2$ from $x=0$ to $x=3$. The area is given by the integral: $$A = \int_{0}^{3} (9 - x^2) \, dx$$

Step 5: Evaluate the integral

$$A = \int_{0}^{3} (9 - x^2) \, dx = \left[9x - \frac{x^3}{3}\right]_{0}^{3}$$ $$A = \left(9(3) - \frac{(3)^3}{3}\right) - \left(9(0) - \frac{(0)^3}{3}\right)$$ $$A = 27 - \frac{27}{3} = 27 - 9 = 18$$

Step 6: State the final answer

The area of the region bounded by $y=9$, $x=0$, and $y=x^2$ is 18 square units.

Final Answer: 18

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Pedagogical Audit

Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the concepts of graphing, integration, and area under a curve to solve the problem. They must interpret the given equations and apply the appropriate integration techniques.

Knowledge Dimension: CONCEPTUAL

Justification: The question requires understanding the concepts of area under a curve, definite integrals, and the relationship between a function and its graph. It involves applying these concepts to a specific problem.

Syllabus Audit: In the context of CBSE Class 12, this is classified as APPLICATION. The question requires the student to apply the concepts of integration to find the area under a curve, which is a standard application of integration in the syllabus.