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

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

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

\int_{0}^{1} (e^{x})^{2} d x

\frac{e ^{2} + 1}{2}

Decimal

4.19452 \dots

Solution steps

\int_{0}^{1} (e^{x})^{2} d x

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

Apply u-substitution

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

= \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

= \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}

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

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

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

Compute the boundaries:

e^{2} + 1

= \frac{1}{2} (e^{2} + 1)

Simplify

= \frac{e ^{2} + 1}{2}

x \to \infty lim (x^{3} - 3 x)

\frac{d}{d x} ((2 x^{2} - 1)^{2})

x \to \infty lim ((2.18 \cdot 1 0^{- 3}) (763) (x) - \frac{1}{2} \cdot x)

\int \frac{1}{e ^{y} - 1} d y

\int 2 x (x^{2} + 3)^{20} d x

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