Questions: Evaluate the iterated integral I = ∫ from 0 to 1 ∫ from 1-y to 1+y (12y^2+6x) dx dy

Solution Steps

To evaluate the iterated integral, we first integrate the inner integral with respect to \(x\) from \(1-y\) to \(1+y\). Then, we integrate the resulting expression with respect to \(y\) from 0 to 1.

We start with the iterated integral

\[ I = \int_{0}^{1} \int_{1-y}^{1+y} (12y^2 + 6x) \, dx \, dy. \]

We first compute the inner integral

\[ \int_{1-y}^{1+y} (12y^2 + 6x) \, dx. \]

Calculating this gives:

\[ \int (12y^2 + 6x) \, dx = 6x^2 + 12y^2 x. \]

Evaluating from \(x = 1-y\) to \(x = 1+y\):

\[ \left[ 6(1+y)^2 + 12y^2(1+y) \right] - \left[ 6(1-y)^2 + 12y^2(1-y) \right]. \]

This simplifies to:

\[ -12y^2(1 - y) + 12y^2(1 + y) - 3(1 - y)^2 + 3(1 + y)^2. \]

Next, we compute the outer integral:

\[ \int_{0}^{1} \left( -12y^2(1 - y) + 12y^2(1 + y) - 3(1 - y)^2 + 3(1 + y)^2 \right) \, dy. \]

After evaluating this integral, we find that

\[ I = 12. \]

Final Answer