Key Concepts and Formulas

• Property of Definite Integrals: The value of a definite integral $\int_a^b f(x) dx$ remains unchanged if we replace $x$ with $a+b-x$. This property is stated as: $\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} }$
• Integral of a Constant: The integral of a constant $c$ with respect to $x$ is $cx$. $\int c \, dx = cx + C$ For definite integrals: $\int_a^b c \, dx = c(b-a)$

Step-by-Step Solution

Step 1: Define the Integral and Identify Limits Let the given integral be $I$. $I = \int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx$ The limits of integration are $a=3$ and $b=6$.

Step 2: Apply the Property of Definite Integrals We use the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. Here, $a+b = 3+6 = 9$. We substitute $x$ with $(9-x)$ in the integrand. Let $f(x) = \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}}$. Then, $f(9-x) = \frac{\sqrt{9-x}}{\sqrt{9-(9-x)} + \sqrt{9-x}} = \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}}$. Applying the property, we get a new expression for $I$: $I = \int\limits_3^6 {{{\sqrt {9 - x} } \over {\sqrt x + \sqrt {9 - x} }}} dx \quad \ldots(2)$

Step 3: Combine the Original and Transformed Integrals We now have two expressions for $I$: $I = \int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx \quad \ldots(1)$ $I = \int\limits_3^6 {{{\sqrt {9 - x} } \over {\sqrt x + \sqrt {9 - x} }}} dx \quad \ldots(2)$ Adding equation (1) and equation (2): $I + I = \int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx + \int\limits_3^6 {{{\sqrt {9 - x} } \over {\sqrt x + \sqrt {9 - x} }}} dx$ Since the limits of integration are the same, we can combine the integrands: $2I = \int\limits_3^6 \left( {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }} + {{\sqrt {9 - x} } \over {\sqrt x + \sqrt {9 - x} }}} \right) dx$ The denominators are identical: $\sqrt{9-x} + \sqrt{x}$. Therefore, we can add the numerators directly: $2I = \int\limits_3^6 \left( {{{\sqrt x + \sqrt {9 - x} } \over {\sqrt {9 - x} + \sqrt x }}} \right) dx$ The term in the parenthesis simplifies to 1: $2I = \int\limits_3^6 1 \, dx$

Step 4: Evaluate the Simplified Integral The integral of $1$ with respect to $x$ is $x$. $2I = \left[ x \right]_3^6$ Applying the limits of integration: $2I = 6 - 3$ $2I = 3$

Step 5: Solve for $I$ Divide both sides by 2 to find the value of the original integral: $I = {3 \over 2}$

Common Mistakes & Tips

• Forgetting the $2I$: A frequent error is to equate the result of the simplified integral directly to $I$ instead of $2I$. Always remember that you are adding the original integral to itself.
• Incorrect Substitution: Ensure that when applying the property $x \to a+b-x$, you correctly substitute into all occurrences of $x$ in the integrand, especially within nested functions like square roots.
• Recognizing Simplification: The power of this property lies in the simplification of the integrand. Look for cases where $f(x) + f(a+b-x)$ results in a constant or a much simpler function, as seen here where it simplified to 1.

Summary

The problem is solved by leveraging a key property of definite integrals: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. By applying this property to the given integral, we obtained an alternative expression for the same integral. Adding the original and the transformed integral resulted in a significant simplification of the integrand to 1, making the final evaluation straightforward. The value of the integral is found to be $\frac{3}{2}$.

The final answer is $\boxed{{3 \over 2}}$.