What is the integral from 0 to 2 of sqrt(e^x) ?

The integral from 0 to 2 of sqrt(e^x) is 2(e-1)

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integral from 0 to 2 of sqrt(e^x)

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

2 (e - 1)

Decimal Notation

3.43656 \dots

Solution steps

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

Apply radical rule:

assuming

a \geq 0

= \int_{0}^{2} e^{\frac{x}{2}} d x

Apply u-substitution

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

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

Take the constant out:

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

= 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)

x^{2} + 1, x + 3, 0

\frac{d ^{2}}{d x ^{2}} (- \frac{2 x}{y})

x \to a lim ([f (x) g (x)])

\frac{( - 2 s + 21 )}{( s ^{2} - 8 s + 20 )}

\frac{d}{d x} (x (x^{3} + 7)^{4})

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