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

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

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

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

1

Solution steps

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

Apply u-substitution

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

= \int_{0}^{1} 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_{0}^{1} e^{u} d u

Apply u-substitution

: \int_{0}^{1} e^{v} \cdot 2 v d v

= \frac{1}{2} \cdot \int_{0}^{1} 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_{0}^{1} e^{v} v d v

Apply Integration By Parts

: [e^{v} v - \int e^{v} d v]_{0}^{1}

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

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

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

Simplify

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

= [e^{v} v - e^{v}]_{0}^{1}

Compute the boundaries:

1

= 1

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

\int_{4}^{5} \frac{4 x ^{2} + x - 2}{x - 3} d x

\int_{0}^{1} \frac{x}{x ^{2} + 3 x + 2} d x

\int_{0}^{1} x^{3} - x d x

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

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