Question:
The value of the integral
\[
\int_a^b \frac{\sqrt{x} \, dx}{\sqrt{x} + \sqrt{a + b - x}}
\]
is:
The value of the integral
\[ \int_a^b \frac{\sqrt{x} \, dx}{\sqrt{x} + \sqrt{a + b - x}} \]
is:
\[ \int_a^b \frac{\sqrt{x} \, dx}{\sqrt{x} + \sqrt{a + b - x}} \]
is:
Show Hint
Use symmetry property:
\[
I = \int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx
\]
Updated On: Mar 23, 2026
- \(\pi\)
- \(\dfrac{1}{2}(b-a)\)
- \(\dfrac{\pi}{2}\)
- b-a
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Replace \(x \to a+b-x\) and add both forms.
Step 2:
\[ I + I = \int_a^b dx = b - a \quad \implies \quad I = \frac{b-a}{2} \]
Step 2:
\[ I + I = \int_a^b dx = b - a \quad \implies \quad I = \frac{b-a}{2} \]
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