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

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

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

\int_{0}^{2} e^{2 x} d x

e^{2} + 1

Decimal

8.38905 \dots

Solution steps

\int_{0}^{2} e^{2 x} d x

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

e^{2} + 1

= e^{2} + 1

\int_{2}^{3} e^{x - 2} - e d x

\int_{0}^{3} t 1 + t^{2} d t

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

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

\int_{1}^{2} (3 x - 1)^{3} d x

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