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

The integral from 0 to 2 of 2e^{-2x} is 1-1/(e^4)

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

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

1 - \frac{1}{e ^{4}}

Decimal

0.98168 \dots

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{0}^{2} e^{- 2 x} d x

Apply u-substitution

: \int_{0}^{- 4} - \frac{1}{2} e^{u} d u

= 2 \cdot \int_{0}^{- 4} - \frac{1}{2} e^{u} d u

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

= 2 (- \int_{- 4}^{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

= 2 (- (- \frac{1}{2} \cdot \int_{- 4}^{0} e^{u} d u))

Use the common integral:

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

= 2 (- (- \frac{1}{2} [e^{u}]_{- 4}^{0}))

Simplify

2 (- (- \frac{1}{2} [e^{u}]_{- 4}^{0})) : [e^{u}]_{- 4}^{0}

= [e^{u}]_{- 4}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{4}}

= 1 - \frac{1}{e ^{4}}

\int_{0}^{\infty} x \cdot e^{- 3 x} d x

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\int_{0}^{2} 2 + x d x

What is the integral from 0 to 2 of 2e^{-2x} ?
The integral from 0 to 2 of 2e^{-2x} is 1-1/(e^4)