What is the integral from 0 to 1 of 5e^{-2sqrt(x)} ?

The integral from 0 to 1 of 5e^{-2sqrt(x)} is -(5(-e^2+3))/(2e^2)

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integral from 0 to 1 of 5e^{-2sqrt(x)}

\int_{0}^{1} 5 e^{- 2 x} d x

- \frac{5 ( - e ^{2} + 3 )}{2 e ^{2}}

Decimal

1.48498 \dots

Solution steps

\int_{0}^{1} 5 e^{- 2 x} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} e^{- 2 x} d x

Apply u-substitution

= 5 \cdot \int_{0}^{1} e^{- 2 u} \cdot 2 u d u

Take the constant out:

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

= 5 \cdot 2 \cdot \int_{0}^{1} e^{- 2 u} u d u

Apply u-substitution

= 5 \cdot 2 \cdot \int_{0}^{- 2} \frac{e ^{v} v}{4} d v

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

= 5 \cdot 2 (- \int_{- 2}^{0} \frac{e ^{v} v}{4} d v)

Take the constant out:

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

= 5 \cdot 2 (- \frac{1}{4} \cdot \int_{- 2}^{0} e^{v} v d v)

Apply Integration By Parts

= 5 \cdot 2 (- \frac{1}{4} [e^{v} v - \int e^{v} d v]_{- 2}^{0})

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

= 5 \cdot 2 (- \frac{1}{4} [e^{v} v - e^{v}]_{- 2}^{0})

Simplify

5 \cdot 2 (- \frac{1}{4} [e^{v} v - e^{v}]_{- 2}^{0}) : - \frac{5}{2} [e^{v} v - e^{v}]_{- 2}^{0}

= - \frac{5}{2} [e^{v} v - e^{v}]_{- 2}^{0}

Compute the boundaries:

- 1 + \frac{3}{e ^{2}}

= - \frac{5}{2} (- 1 + \frac{3}{e ^{2}})

Simplify

= - \frac{5 ( - e ^{2} + 3 )}{2 e ^{2}}

What is the integral from 0 to 1 of 5e^{-2sqrt(x)} ?
The integral from 0 to 1 of 5e^{-2sqrt(x)} is -(5(-e^2+3))/(2e^2)