What is the integral from 0 to 2 of 8e^{2t} ?

The integral from 0 to 2 of 8e^{2t} is 4(e^4-1)

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integral from 0 to 2 of 8e^{2t}

\int_{0}^{2} 8 e^{2 t} d t

4 (e^{4} - 1)

Decimal

214.39260 \dots

Solution steps

\int_{0}^{2} 8 e^{2 t} d t

Take the constant out:

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

= 8 \cdot \int_{0}^{2} e^{2 t} d t

Apply u-substitution

: \int_{0}^{4} e^{u} \frac{1}{2} d u

= 8 \cdot \int_{0}^{4} 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

= 8 \cdot \frac{1}{2} \cdot \int_{0}^{4} e^{u} d u

Use the common integral:

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

= 8 \cdot \frac{1}{2} [e^{u}]_{0}^{4}

Simplify

8 \cdot \frac{1}{2} [e^{u}]_{0}^{4} : 4 [e^{u}]_{0}^{4}

= 4 [e^{u}]_{0}^{4}

Compute the boundaries:

e^{4} - 1

= 4 (e^{4} - 1)

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What is the integral from 0 to 2 of 8e^{2t} ?
The integral from 0 to 2 of 8e^{2t} is 4(e^4-1)