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- \int_{0}^{\infty} e^{- 4 t}\, dt
First find an antiderivative F, then evaluate F(b) − F(a).
- u = - 4 t,\quad du = -4\, dt
Substitute u = - 4 t.
- \int e^{- 4 t}\, dt = \int - \frac{e^{u}}{4}\, d_u
Rewrite the integral in terms of u.
- \int - \frac{e^{u}}{4}\, d_u = - \frac{1}{4} \int e^{u}\, d_u
Pull the constant - \frac{1}{4} out of the integral.
- \int e^{u}\, d_u = e^{u}
∫ aᵘ du = aᵘ / ln a (for eˣ that is just eˣ).
- = - \frac{e^{- 4 t}}{4}
Substitute back u = - 4 t.
- F(\infty) - F(0) = \left(0\right) - \left(- \frac{1}{4}\right)
Fundamental theorem of calculus: plug in the limits.
- = \frac{1}{4}
Simplify.
Reveal the answer
\frac{1}{4}