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

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

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

\int_{- 1}^{0} e^{x + 1} d x

2

Solution steps

\int_{- 1}^{0} e^{x + 1} d x

Apply u-substitution

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

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

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{1}

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

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

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

Compute the boundaries:

1

= 2 \cdot 1

Simplify

= 2

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