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

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

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

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

2

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

= 2 [e^{u} u - \int e^{u} d u]_{0}^{1}

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

= 2 [e^{u} u - e^{u}]_{0}^{1}

Compute the boundaries:

1

= 2 \cdot 1

Simplify

= 2

\frac{d}{d x} (x + a)

\frac{d}{d x} (ln (x - y))

\frac{\partial}{\partial x} (\frac{x \cdot y}{x ^{2} + y ^{2}})

\int x (sin (x) - cos (x^{2})) d x

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