CUET · MATHS · PYQ PAPER 2025
The value of the definite integral \(I=\int_0^2 x \sqrt{2-x} d x\) is :
- A \(\frac{16}{15 \sqrt{2}}\)
- B \(\frac{16 \sqrt{3}}{15}\)
- C \(\frac{16 \sqrt{2}}{15}\)
- D \(\frac{5 \sqrt{2}}{7}\)
Answer & Solution
Correct Answer
(C) \(\frac{16 \sqrt{2}}{15}\)
Step-by-step Solution
Detailed explanation
\(I=\int_0^2 x \sqrt{2-x} d x\) Let \(u=2-x \Rightarrow x=2-u, du=-dx\). Limits: \(x=0 \Rightarrow u=2\), \(x=2 \Rightarrow u=0\). \(I=\int_2^0 (2-u) \sqrt{u} (-du) = \int_0^2 (2u^{1/2}-u^{3/2}) du\)…
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