KCET · Maths · Definite Integration
The value of \(\int_{0}^{4042} \frac{\sqrt{x} d x}{\sqrt{x}+\sqrt{4042-x}}\) is equal to
- A 4042
- B 2021
- C 8084
- D 1010
Answer & Solution
Correct Answer
(B) 2021
Step-by-step Solution
Detailed explanation
Let \(I=\int_{0}^{4042} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4042-x}} d x...(i)\)
Using the property, \(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\)
\(I=\int_{0}^{4042} \frac{\sqrt{4042-x}}{\sqrt{4042-x}+\sqrt{x}}...(ii)\)
Adding Eqs. (i) and (ii),
\(2 I=\int_{0}^{4042} \frac{\sqrt{x}+\sqrt{4042-x}}{\sqrt{4042-x}+\sqrt{x}} d x\)
\(\Rightarrow \quad 2 I=\int_{0}^{4042} 1 d x\)
\(\Rightarrow \quad 2 I=4042\)
\(\Rightarrow \quad I=2021\)
Using the property, \(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\)
\(I=\int_{0}^{4042} \frac{\sqrt{4042-x}}{\sqrt{4042-x}+\sqrt{x}}...(ii)\)
Adding Eqs. (i) and (ii),
\(2 I=\int_{0}^{4042} \frac{\sqrt{x}+\sqrt{4042-x}}{\sqrt{4042-x}+\sqrt{x}} d x\)
\(\Rightarrow \quad 2 I=\int_{0}^{4042} 1 d x\)
\(\Rightarrow \quad 2 I=4042\)
\(\Rightarrow \quad I=2021\)
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