What is the integral from 0 to 10 of 2e^{-2x} ?

The integral from 0 to 10 of 2e^{-2x} is 1-1/(e^{20)}

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integral from 0 to 10 of 2e^{-2x}

\int_{0}^{10} 2 e^{- 2 x} d x

1 - \frac{1}{e ^{20}}

Solution steps

\int_{0}^{10} 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}^{10} e^{- 2 x} d x

Apply u-substitution

: \int_{0}^{- 20} - \frac{1}{2} e^{u} d u

= 2 \cdot \int_{0}^{- 20} - \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_{- 20}^{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_{- 20}^{0} e^{u} d u))

Use the common integral:

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

= 2 (- (- \frac{1}{2} [e^{u}]_{- 20}^{0}))

Simplify

2 (- (- \frac{1}{2} [e^{u}]_{- 20}^{0})) : [e^{u}]_{- 20}^{0}

= [e^{u}]_{- 20}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{20}}

= 1 - \frac{1}{e ^{20}}

\int_{0}^{2} 20 t^{2} e^{- t} d t

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

\int_{0}^{1} 4 x^{3} 1 + 9 x^{4} d x

\int_{0}^{a} cosh^{2} (\frac{t}{a}) d t

\int_{1}^{b} \frac{1}{x ^{p}} d x

What is the integral from 0 to 10 of 2e^{-2x} ?
The integral from 0 to 10 of 2e^{-2x} is 1-1/(e^{20)}