What is the integral from 0 to 1 of 8e^{2sqrt(x)} ?

The integral from 0 to 1 of 8e^{2sqrt(x)} is 4(e^2+1)

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integral from 0 to 1 of 8e^{2sqrt(x)}

\int_{0}^{1} 8 e^{2 x} d x

4 (e^{2} + 1)

Decimal Notation

33.55622 \dots

Solution steps

\int_{0}^{1} 8 e^{2 x} d x

Take the constant out:

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

= 8 \cdot \int_{0}^{1} e^{2 x} d x

Apply u-substitution

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

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

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{2}

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

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

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

Simplify

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

= 4 [e^{u} u - e^{u}]_{0}^{2}

Compute the boundaries:

e^{2} + 1

= 4 (e^{2} + 1)

\frac{d}{d x} (ln (x^{2} e^{- x}))

x y - 2 y^{2} = 5, (11, 5)

\int_{1}^{2} \frac{9}{x ^{3} + 4 x} d x

\frac{d}{d x} (\frac{5}{2} x^{2})

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