KCET · Maths · Definite Integration
The value of \( \int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x \) is
- A \( 10 \)
- B \( 00 \)
- C \( 08 \)
- D \( 13 \)
Answer & Solution
Correct Answer
(D) \( 13 \)
Step-by-step Solution
Detailed explanation
Given that, \( I=\int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x \rightarrow(1) \)
We know that, \( \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \)
So, \( I=\int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x \rightarrow(2) \)
Adding Eqs. (1) and (2), we get
\( 2 I=\int_{2}^{8} d x=[x]_{2}^{8} \) \( \Rightarrow 2 I=6 \Rightarrow I=3 \)
We know that, \( \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \)
So, \( I=\int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x \rightarrow(2) \)
Adding Eqs. (1) and (2), we get
\( 2 I=\int_{2}^{8} d x=[x]_{2}^{8} \) \( \Rightarrow 2 I=6 \Rightarrow I=3 \)
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