∫_0^2▒〖x√(2x-1) dx〗

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on Dec 27, 2024 at 1:40 am



You asked:

Evaluate the integral: ∫02x2x−1dx\int_{0}^{2}{x \sqrt{2 x - 1}}dx

MathBot Answer:

Definite integral

∫02x2x−1dx=i15+735\int_{0}^{2}{x \sqrt{2 x - 1}}dx = \frac{i}{15} + \frac{7 \sqrt{3}}{5}

ii is the imaginary unit, defined as i2=−1i^2 = -1.

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