∫_0^2▒〖x√(2x-1) dx〗

asked by guest
on Dec 27, 2024 at 1:40 am



You asked:

Evaluate the integral: \(\int_{0}^{2}{x \sqrt{2 x - 1}}dx\)

MathBot Answer:

Definite integral

\[\int_{0}^{2}{x \sqrt{2 x - 1}}dx = \frac{i}{15} + \frac{7 \sqrt{3}}{5}\]

\(i\) is the imaginary unit, defined as \(i^2 = -1\).

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