What is the integral from 0 to 1 of e^{-2x} ?

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

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integral from 0 to 1 of e^{-2x}

\int_{0}^{1} e^{- 2 x} d x

\frac{e ^{2} - 1}{2 e ^{2}}

Decimal Notation

0.43233 \dots

Solution steps

\int_{0}^{1} e^{- 2 x} d x

Apply u-substitution

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

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

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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

1 - \frac{1}{e ^{2}}

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

Simplify

= \frac{e ^{2} - 1}{2 e ^{2}}

y = x, x = 4

\int tan^{7} (x) d x

y = (x^{5} + 1)^{6}

y = 4 x^{2} + 2

\int t^{4} + 2 t^{2} + 1 d t

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