What is the integral from 0 to t of e^{-2(t-x)} ?

The integral from 0 to t of e^{-2(t-x)} is 1/2 (1-e^{-2t})

Popular >

integral from 0 to t of e^{-2(t-x)}

\int_{0}^{t} e^{- 2 (t - x)} d x

\frac{1}{2} (1 - e^{- 2 t})

Solution steps

\int_{0}^{t} e^{- 2 (t - x)} d x

Apply u-substitution

: \int_{- 2 t}^{0} e^{u} \frac{1}{2} d u

= \int_{- 2 t}^{0} e^{u} \frac{1}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{- 2 t}^{0} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{2} [e^{u}]_{- 2 t}^{0}

Compute the boundaries:

1 - e^{- 2 t}

= \frac{1}{2} (1 - e^{- 2 t})

\int_{\frac{π}{4}}^{\frac{π}{2}} csc^{4} (x) d x

\int_{1}^{3} y d y

\int_{0}^{\frac{π}{2}} 8 cos (t) sin (t) d t

\int_{- 3}^{3} x^{4} - x^{2} + 20 d x

\int_{- 1}^{1} 3 x^{2} (x^{3} + 2)^{2} d x

What is the integral from 0 to t of e^{-2(t-x)} ?
The integral from 0 to t of e^{-2(t-x)} is 1/2 (1-e^{-2t})