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

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

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

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

2 e^{3}

Decimal Notation

40.17107 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{9} e^{u} \frac{1}{2} d u

= \int_{1}^{9} e^{u} \frac{1}{2} 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_{1}^{9} e^{u} d u

Apply u-substitution

: \int_{1}^{3} e^{v} \cdot 2 v d v

= \frac{1}{2} \cdot \int_{1}^{3} e^{v} \cdot 2 v d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int_{1}^{3} e^{v} v d v

Apply Integration By Parts

: [e^{v} v - \int e^{v} d v]_{1}^{3}

= \frac{1}{2} \cdot 2 [e^{v} v - \int e^{v} d v]_{1}^{3}

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

= \frac{1}{2} \cdot 2 [e^{v} v - e^{v}]_{1}^{3}

Simplify

\frac{1}{2} \cdot 2 [e^{v} v - e^{v}]_{1}^{3} : [e^{v} v - e^{v}]_{1}^{3}

= [e^{v} v - e^{v}]_{1}^{3}

Compute the boundaries:

2 e^{3}

= 2 e^{3}

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