What is the integral from 0 to 9 of e^{sqrt(x)} ?

The integral from 0 to 9 of e^{sqrt(x)} is 2(2e^3+1)

Popular >

integral from 0 to 9 of e^{sqrt(x)}

\int_{0}^{9} e^{x} d x

2 (2 e^{3} + 1)

Decimal

82.34214 \dots

Solution steps

\int_{0}^{9} e^{x} d x

Apply u-substitution

: \int_{0}^{3} e^{u} \cdot 2 u d u

= \int_{0}^{3} 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}^{3} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{3}

= 2 [e^{u} u - \int e^{u} d u]_{0}^{3}

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

= 2 [e^{u} u - e^{u}]_{0}^{3}

Compute the boundaries:

2 e^{3} + 1

= 2 (2 e^{3} + 1)

\int_{0}^{1} x^{2} - 3 x + 8 d x

\int_{6}^{6} 1

\int_{0}^{2} 17 x (x^{2} + 1)^{2} d x

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

\int_{- 5}^{5} (7 + 25 - x^{2}) d x

What is the integral from 0 to 9 of e^{sqrt(x)} ?
The integral from 0 to 9 of e^{sqrt(x)} is 2(2e^3+1)