What is the integral from 0 to 4 of 6e^{2t} ?

The integral from 0 to 4 of 6e^{2t} is 3(e^8-1)

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integral from 0 to 4 of 6e^{2t}

\int_{0}^{4} 6 e^{2 t} d t

3 (e^{8} - 1)

Decimal Notation

8939.87396 \dots

Solution steps

\int_{0}^{4} 6 e^{2 t} d t

Take the constant out:

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

= 6 \cdot \int_{0}^{4} e^{2 t} d t

Apply u-substitution

: \int_{0}^{8} e^{u} \frac{1}{2} d u

= 6 \cdot \int_{0}^{8} 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

= 6 \cdot \frac{1}{2} \cdot \int_{0}^{8} e^{u} d u

Use the common integral:

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

= 6 \cdot \frac{1}{2} [e^{u}]_{0}^{8}

Simplify

6 \cdot \frac{1}{2} [e^{u}]_{0}^{8} : 3 [e^{u}]_{0}^{8}

= 3 [e^{u}]_{0}^{8}

Compute the boundaries:

e^{8} - 1

= 3 (e^{8} - 1)

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What is the integral from 0 to 4 of 6e^{2t} ?
The integral from 0 to 4 of 6e^{2t} is 3(e^8-1)