silemus.

34Improper integral of e−2xe^{-2x}

Evaluate ∫0∞e−2x dx\displaystyle \int_0^{\infty} e^{-2x}\,dx.

Turn the infinite bound into a limit as t→∞. The antiderivative of e^(-2x) is -½e^(-2x); plugging in bounds and letting t→∞ makes the e^(-2t) term vanish, leaving ½. So the integral equals 1/2.

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