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resolver paso a paso la integral definida

To solve the given definite integral step by step, we have:

\[
\int_{0}^{a} \left(\sqrt{a} - \sqrt{x}\right)^2 \, dx
\]

### Step 1: Expand the integrand
First, let's expand the square in the integrand:

\[
\left(\sqrt{a} - \sqrt{x}\right)^2 = \left(\sqrt{a} - \sqrt{x}\right) \times \left(\sqrt{a} - \sqrt{x}\right)
\]

Using the distributive property (also known as FOIL in binomials), we get:

\[
\left(\sqrt{a} - \sqrt{x}\right)^2 = (\sqrt{a})^2 - 2\sqrt{a}\sqrt{x} + (\sqrt{x})^2
\]

Simplify each term:

\[
= a - 2\sqrt{a}\sqrt{x} + x
\]

So the integral becomes:

\[
\int_{0}^{a} \left(a - 2\sqrt{a}\sqrt{x} + x\right) \, dx
\]

### Step 2: Integrate each term separately
Now we can integrate each term individually:

1. The integral of \(a\) with respect to \(x\):

\[
\int_{0}^{a} a \, dx = a \int_{0}^{a} 1 \, dx = a \cdot \left[x\right]_{0}^{a} = a \cdot (a - 0) = a^2
\]

2. The integral of \(-2\sqrt{a}\sqrt{x}\) with respect to \(x\):

\[
\int_{0}^{a} -2\sqrt{a}\sqrt{x} \, dx = -2\sqrt{a} \int_{0}^{a} \sqrt{x} \, dx
\]

To integrate \(\sqrt{x}\), recall that \(\sqrt{x} = x^{1/2}\):

\[
\int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}
\]

Thus,

\[
\int_{0}^{a} \sqrt{x} \, dx = \frac{2}{3} \left[ x^{3/2} \right]_{0}^{a} = \frac{2}{3} \left( a^{3/2} - 0^{3/2} \right) = \frac{2}{3} a^{3/2}
\]

So, the integral becomes:

\[
-2\sqrt{a} \cdot \frac{2}{3} a^{3/2} = -\frac{4}{3} a^{2}
\]

3. The integral of \(x\) with respect to \(x\):

\[
\int_{0}^{a} x \, dx = \frac{1}{2} x^2 \Big|_0^a = \frac{1}{2} a^2
\]

### Step 3: Combine the results
Now, add up the integrals:

\[
a^2 - \frac{4}{3} a^2 + \frac{1}{2} a^2
\]

Combine like terms:

\[
= \left(1 - \frac{4}{3} + \frac{1}{2}\right) a^2
\]

Finding a common denominator:

\[
= \left(\frac{6}{6} - \frac{8}{6} + \frac{3}{6}\right) a^2 = \frac{1}{6} a^2
\]

### Final Answer
Thus, the value of the definite integral is:

\[
\int_{0}^{a} \left(\sqrt{a} - \sqrt{x}\right)^2 \, dx = \frac{1}{6} a^2
\]

Would you like any further explanation or details?

Here are some related questions:
1. How would you handle an integral with a higher degree polynomial instead of a square?
2. What happens if the limits of integration are different?
3. How does the presence of a constant inside the square affect the integral?
4. Can you solve a similar integral involving a trigonometric function?
5. How can definite integrals be applied in real-world situations?

**Tip:** Always expand and simplify the integrand before attempting to integrate, as it can significantly simplify the problem.

Learn how to solve the definite integral (sqrt(a) - sqrt(x))^2 from 0 to a with detailed step-by-step instructions. This problem covers integration techniques, polynomial expansion, and the properties of definite integrals. Suitable for advanced undergraduate students.

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