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

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

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

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

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

Decimal Notation

0.86466 \dots

Solution steps

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

Apply u-substitution

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

= 2 \cdot \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

= 2 (- \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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

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

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

- sin (x) + cos (x) + x - \frac{1}{x ^{2}}

\frac{d}{d x} (x^{3} - 6 x^{2} + 8 x)

y^{^{''}} + 4 y = 8 x^{2}

x^{3} + y^{3} = 9

\int (7 x^{2}) e x^{3} + 1 d x

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