To solve and verify the given double integral:

$\int_0^1 \int_0^{1-x} (x^2 + y^2) \, dy \, dx,$

and check if it equals $\frac{1}{6}$, let's proceed step by step:

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Step 1: Rewrite the Double Integral

The integral is: $\int_0^1 \int_0^{1-x} (x^2 + y^2) \, dy \, dx.$

The limits for $y$ are from 0 to $1-x$, and for $x$, they are from 0 to 1.

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Step 2: Integrate with Respect to $y$

For the inner integral: $\int_0^{1-x} (x^2 + y^2) \, dy,$

we separate $x^2$ and $y^2$ for easier integration: $\int_0^{1-x} x^2 \, dy + \int_0^{1-x} y^2 \, dy.$

First Term:

$\int_0^{1-x} x^2 \, dy = x^2 \int_0^{1-x} 1 \, dy = x^2 [y]_0^{1-x} = x^2 (1-x).$

Second Term:

$\int_0^{1-x} y^2 \, dy = \left[\frac{y^3}{3}\right]_0^{1-x} = \frac{(1-x)^3}{3}.$

Combining both terms: $\int_0^{1-x} (x^2 + y^2) \, dy = x^2 (1-x) + \frac{(1-x)^3}{3}.$

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Step 3: Integrate with Respect to $x$

Now, the outer integral becomes: $\int_0^1 \left[ x^2 (1-x) + \frac{(1-x)^3}{3} \right] dx.$

Separate into two integrals: $\int_0^1 x^2 (1-x) \, dx + \int_0^1 \frac{(1-x)^3}{3} \, dx.$

First Term:

$\int_0^1 x^2 (1-x) \, dx = \int_0^1 (x^2 - x^3) \, dx = \int_0^1 x^2 \, dx - \int_0^1 x^3 \, dx.$

$\int_0^1 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}.$ $\int_0^1 x^3 \, dx = \left[\frac{x^4}{4}\right]_0^1 = \frac{1}{4}.$

$\int_0^1 x^2 (1-x) \, dx = \frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12}.$

Second Term:

$\int_0^1 \frac{(1-x)^3}{3} \, dx = \frac{1}{3} \int_0^1 (1-x)^3 \, dx.$

Let $u = 1-x$, so $du = -dx$ and when $x = 0, u = 1$; when $x = 1, u = 0$.

$\int_0^1 (1-x)^3 \, dx = \int_1^0 u^3 (-du) = \int_0^1 u^3 \, du.$

$\int_0^1 u^3 \, du = \left[\frac{u^4}{4}\right]_0^1 = \frac{1}{4}.$

Thus: $\int_0^1 \frac{(1-x)^3}{3} \, dx = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}.$

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Step 4: Combine Results

Now sum the two parts: $\int_0^1 \left[ x^2 (1-x) + \frac{(1-x)^3}{3} \right] dx = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}.$

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Verification

The calculated result is: $\boxed{\frac{1}{6}},$ which matches the given value. The integral is verified.

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Would you like further clarification or a different example?

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Related Questions:

1. How does changing the bounds of integration affect the result?
2. Can we solve this using a numerical integration method?
3. What are other real-world applications of double integrals?
4. How do symmetry properties simplify double integrals?
5. Can this problem be extended to triple integrals?

Tip:

Always sketch the integration region to confirm bounds and simplify integration.