What is the integral from 0 to 2 of e^{0.5x} ?

The integral from 0 to 2 of e^{0.5x} is 2(e-1)

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integral from 0 to 2 of e^{0.5x}

\int_{0}^{2} e^{0.5 x} d x

2 (e - 1)

Decimal

3.43656 \dots

Solution steps

\int_{0}^{2} e^{0.5 x} d x

Apply u-substitution

: \int_{0}^{1} e^{u} \frac{1}{0.5} d u

= \int_{0}^{1} e^{u} \frac{1}{0.5} d u

Take the constant out:

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

= \frac{1}{0.5} \cdot \int_{0}^{1} e^{u} d u

Simplify

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

Use the common integral:

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

= 2 [e^{u}]_{0}^{1}

Compute the boundaries:

e - 1

= 2 (e - 1)

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What is the integral from 0 to 2 of e^{0.5x} ?
The integral from 0 to 2 of e^{0.5x} is 2(e-1)