Evaluate the definite integral from 0 to 2 of √(x) / (√(x) + √(g - x)) dx.
Understand the Problem
The question is asking to evaluate a definite integral involving a square root function. The approach involves applying properties of definite integrals and performing algebraic manipulations to derive the result.
Answer
The value of the definite integral is $1$.
Answer for screen readers
The value of the definite integral is
$$ I = 1. $$
Steps to Solve
- Set up the equation
We want to evaluate the definite integral:
$$ I = \int_0^2 \frac{\sqrt{x}}{\sqrt{x} + \sqrt{g - x}} , dx. $$
- Apply the substitution
Using the substitution ( x = g - u ), the differential ( dx = -du ). The limits change from ( x = 0 ) to ( u = g ) and from ( x = 2 ) to ( u = g - 2 ):
$$ I = \int_g^{g-2} \frac{\sqrt{g - u}}{\sqrt{g - u} + \sqrt{u}} (-du) = \int_{g-2}^g \frac{\sqrt{g - u}}{\sqrt{g - u} + \sqrt{u}} , du. $$
- Add the two integrals
Now we have two expressions for ( I ):
$$ I = \int_0^2 \frac{\sqrt{x}}{\sqrt{x} + \sqrt{g - x}} , dx $$
and
$$ I = \int_{g-2}^g \frac{\sqrt{g - u}}{\sqrt{g - u} + \sqrt{u}} , du. $$
Adding these gives: $$ 2I = \int_0^2 \frac{\sqrt{x}}{\sqrt{x} + \sqrt{g - x}} , dx + \int_0^2 \frac{\sqrt{g - x}}{\sqrt{g - x} + \sqrt{x}} , dx. $$
- Combine the integrals
Adding the expressions inside the integral gives:
$$ 2I = \int_0^2 1 , dx = 2. $$
- Solve for ( I )
Now, divide both sides by 2:
$$ I = \frac{2}{2} = 1. $$
The value of the definite integral is
$$ I = 1. $$
More Information
This result shows how symmetry can be leveraged in integrals involving square roots. The function's behavior around the midpoint of the interval often allows simplification.
Tips
- Forgetting to change the limits when performing substitutions can lead to incorrect evaluations.
- Doing algebraic manipulations incorrectly, especially when combining integrals.