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Calculus · Definite-Integrals

Integrate e^(-2t) from 0 to oo

\int_{0}^{\infty} e^{- 2 t}\, dt

Step by step

  1. \int_{0}^{\infty} e^{- 2 t}\, dt

    First find an antiderivative F, then evaluate F(b) − F(a).

  2. u = - 2 t,\quad du = -2\, dt

    Substitute u = - 2 t.

  3. \int e^{- 2 t}\, dt = \int - \frac{e^{u}}{2}\, d_u

    Rewrite the integral in terms of u.

  4. \int - \frac{e^{u}}{2}\, d_u = - \frac{1}{2} \int e^{u}\, d_u

    Pull the constant - \frac{1}{2} out of the integral.

  5. \int e^{u}\, d_u = e^{u}

    ∫ aᵘ du = aᵘ / ln a (for eˣ that is just eˣ).

  6. = - \frac{e^{- 2 t}}{2}

    Substitute back u = - 2 t.

  7. F(\infty) - F(0) = \left(0\right) - \left(- \frac{1}{2}\right)

    Fundamental theorem of calculus: plug in the limits.

  8. = \frac{1}{2}

    Simplify.

Reveal the answer
\frac{1}{2}