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

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

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integral from 0 to 1 of e^{1-t}

\int_{0}^{1} e^{1 - t} d t

e - 1

Decimal

1.71828 \dots

Solution steps

\int_{0}^{1} e^{1 - t} d t

Apply u-substitution

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

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

= - (- \int_{0}^{1} e^{u} d u)

Use the common integral:

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

= - (- [e^{u}]_{0}^{1})

Simplify

= [e^{u}]_{0}^{1}

Compute the boundaries:

e - 1

= e - 1

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