What is the integral from 0 to t of e^{2u} ?

The integral from 0 to t of e^{2u} is 1/2 (e^{2t}-1)

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integral from 0 to t of e^{2u}

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

\frac{1}{2} (e^{2 t} - 1)

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{2 t} - 1

= \frac{1}{2} (e^{2 t} - 1)

What is the integral from 0 to t of e^{2u} ?
The integral from 0 to t of e^{2u} is 1/2 (e^{2t}-1)