What is the integral from 0 to t of e^{t-v} ?

The integral from 0 to t of e^{t-v} is -1+e^t

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integral from 0 to t of e^{t-v}

\int_{0}^{t} e^{t - v} d v

- 1 + e^{t}

Solution steps

\int_{0}^{t} e^{t - v} d v

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= - [e^{u}]_{t}^{0}

Compute the boundaries:

1 - e^{t}

= - (1 - e^{t})

Simplify

= - 1 + e^{t}

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What is the integral from 0 to t of e^{t-v} ?
The integral from 0 to t of e^{t-v} is -1+e^t