Class CBSE Class 12 Mathematics Integrals Q #899
KNOWLEDGE BASED
APPLY
3 Marks 2023 SA
27. (b) OR: Evaluate: $\int_{-2}^{2}\frac{x^{2}}{1+5^{x}}dx$
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Step-by-Step Solution

  1. Let $I = \int_{-2}^{2} \frac{x^2}{1+5^x} dx$ ...(1)
  2. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$, we have: $I = \int_{-2}^{2} \frac{(-x)^2}{1+5^{-x}} dx = \int_{-2}^{2} \frac{x^2}{1+5^{-x}} dx = \int_{-2}^{2} \frac{x^2}{\frac{5^x+1}{5^x}} dx = \int_{-2}^{2} \frac{x^2 5^x}{1+5^x} dx$ ...(2)
  3. Adding equations (1) and (2): $2I = \int_{-2}^{2} \frac{x^2}{1+5^x} dx + \int_{-2}^{2} \frac{x^2 5^x}{1+5^x} dx = \int_{-2}^{2} \frac{x^2 + x^2 5^x}{1+5^x} dx = \int_{-2}^{2} \frac{x^2(1+5^x)}{1+5^x} dx = \int_{-2}^{2} x^2 dx$
  4. $2I = \int_{-2}^{2} x^2 dx = \left[ \frac{x^3}{3} \right]_{-2}^{2} = \frac{2^3}{3} - \frac{(-2)^3}{3} = \frac{8}{3} - \frac{-8}{3} = \frac{8}{3} + \frac{8}{3} = \frac{16}{3}$
  5. $2I = \frac{16}{3}$, so $I = \frac{16}{3 \times 2} = \frac{8}{3}$

Correct Answer: 8/3

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the properties of definite integrals to solve the problem. Specifically, they need to use the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ to simplify the integral and then evaluate it.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure involving the properties of definite integrals to arrive at the solution. This involves applying the correct property, simplifying the integral, and then evaluating it.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge and application of definite integrals, a standard topic in the syllabus.

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