Integral ∫_0^1 ∫_x^1 (x+y) dy dx=⋯

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Hitunglah \( \displaystyle \int_0^1 \int_x^1 \ (x+y) \ dy \ dx \).

Pembahasan:

\begin{aligned} \int_0^1 \int_x^1 \ (x+y) \ dy \ dx &= \int_0^1 \left[ xy+\frac{1}{2}y^2 \right]_x^1 \ dx \\[8pt] &= \int_0^1 \ \left[ \left( x + \frac{1}{2} \right)-\left(x^2+\frac{1}{2}x^2 \right) \right] \ dx \\[8pt] &= \int_0^1 \ \left( x - \frac{3}{2}x^2 + \frac{1}{2} \right) \ dx \\[8pt] &= \left[ \frac{1}{2}x^2 - \frac{1}{2}x^3 + \frac{1}{2}x \right]_0^1 \\[8pt] &= \left( \frac{1}{2} - \frac{1}{2} + \frac{1}{2} \right)-0 \\[8pt] &= \frac{1}{2} \end{aligned}