What is the integral from 0 to 2 of 64e^{2x} ?

The integral from 0 to 2 of 64e^{2x} is 32(e^4-1)

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integral from 0 to 2 of 64e^{2x}

\int_{0}^{2} 64 e^{2 x} d x

32 (e^{4} - 1)

Decimal

1715.14080 \dots

Solution steps

\int_{0}^{2} 64 e^{2 x} d x

Take the constant out:

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

= 64 \cdot \int_{0}^{2} e^{2 x} d x

Apply u-substitution

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

= 64 \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

= 64 \cdot \frac{1}{2} \cdot \int_{0}^{4} e^{u} d u

Use the common integral:

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

= 64 \cdot \frac{1}{2} [e^{u}]_{0}^{4}

Simplify

64 \cdot \frac{1}{2} [e^{u}]_{0}^{4} : 32 [e^{u}]_{0}^{4}

= 32 [e^{u}]_{0}^{4}

Compute the boundaries:

e^{4} - 1

= 32 (e^{4} - 1)

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What is the integral from 0 to 2 of 64e^{2x} ?
The integral from 0 to 2 of 64e^{2x} is 32(e^4-1)