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

The integral from 0 to t of 2e^{2x} is e^{2t}-1

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

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

e^{2 t} - 1

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{0}^{t} e^{2 x} d x

Apply u-substitution

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

= 2 \cdot \int_{0}^{2 t} 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

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

Use the common integral:

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

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

Simplify

2 \cdot \frac{1}{2} [e^{u}]_{0}^{2 t} : [e^{u}]_{0}^{2 t}

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

Compute the boundaries:

e^{2 t} - 1

= e^{2 t} - 1

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What is the integral from 0 to t of 2e^{2x} ?
The integral from 0 to t of 2e^{2x} is e^{2t}-1