What is the integral from 0 to 4 of e^{sqrt(x)} ?

The integral from 0 to 4 of e^{sqrt(x)} is 2(e^2+1)

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integral from 0 to 4 of e^{sqrt(x)}

\int_{0}^{4} e^{x} d x

2 (e^{2} + 1)

Decimal Notation

16.77811 \dots

Solution steps

\int_{0}^{4} e^{x} d x

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

e^{2} + 1

= 2 (e^{2} + 1)

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\int (8 x + 9) d x

What is the integral from 0 to 4 of e^{sqrt(x)} ?
The integral from 0 to 4 of e^{sqrt(x)} is 2(e^2+1)