To solve the given double integral, we can first focus on the inner integral with respect to \( y \), treating \( x \) as a constant. After evaluating the inner integral, we then proceed to evaluate the outer integral with respect to \( x \).
To solve the double integral \(\int_{0}^{1} \int_{0}^{1} x(y + x^2)^4 \, dy \, dx\), we first evaluate the inner integral with respect to \( y \). The expression to integrate is \( x(y + x^2)^4 \).
\[
\int_{0}^{1} x(y + x^2)^4 \, dy = x \left[ \frac{(y + x^2)^5}{5} \right]_{0}^{1}
\]
Evaluating this from \( y = 0 \) to \( y = 1 \), we get:
\[
x \left( \frac{(1 + x^2)^5}{5} - \frac{(x^2)^5}{5} \right) = x \left( \frac{1 + 5x^2 + 10x^4 + 10x^6 + 5x^8 + x^{10} - x^{10}}{5} \right)
\]
Simplifying, the inner integral becomes:
\[
x \left( \frac{1 + 5x^2 + 10x^4 + 10x^6 + 5x^8}{5} \right) = \frac{x + 5x^3 + 10x^5 + 10x^7 + 5x^9}{5}
\]
Next, we evaluate the outer integral with respect to \( x \):
\[
\int_{0}^{1} \left( \frac{x + 5x^3 + 10x^5 + 10x^7 + 5x^9}{5} \right) \, dx
\]
This simplifies to:
\[
\frac{1}{5} \left( \int_{0}^{1} x \, dx + 5 \int_{0}^{1} x^3 \, dx + 10 \int_{0}^{1} x^5 \, dx + 10 \int_{0}^{1} x^7 \, dx + 5 \int_{0}^{1} x^9 \, dx \right)
\]
Calculating each integral separately:
\[
\int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2}
\]
\[
\int_{0}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{4}
\]
\[
\int_{0}^{1} x^5 \, dx = \left[ \frac{x^6}{6} \right]_{0}^{1} = \frac{1}{6}
\]
\[
\int_{0}^{1} x^7 \, dx = \left[ \frac{x^8}{8} \right]_{0}^{1} = \frac{1}{8}
\]
\[
\int_{0}^{1} x^9 \, dx = \left[ \frac{x^{10}}{10} \right]_{0}^{1} = \frac{1}{10}
\]
Substituting these results back, we have:
\[
\frac{1}{5} \left( \frac{1}{2} + 5 \times \frac{1}{4} + 10 \times \frac{1}{6} + 10 \times \frac{1}{8} + 5 \times \frac{1}{10} \right)
\]
Simplifying further:
\[
\frac{1}{5} \left( \frac{1}{2} + \frac{5}{4} + \frac{10}{6} + \frac{10}{8} + \frac{5}{10} \right) = \frac{1}{5} \left( \frac{15}{30} + \frac{37.5}{30} + \frac{50}{30} + \frac{37.5}{30} + \frac{15}{30} \right)
\]
\[
= \frac{1}{5} \left( \frac{155}{30} \right) = \frac{31}{30}
\]
The value of the double integral is \(\boxed{\frac{31}{30}}\).
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