What is the integral from 1 to 9 of e^{sqrt(x)} ?

The integral from 1 to 9 of e^{sqrt(x)} is 4e^3

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integral from 1 to 9 of e^{sqrt(x)}

\int_{1}^{9} e^{x} d x

4 e^{3}

Solution steps

\int_{1}^{9} e^{x} d x

Apply u-substitution

: \int_{1}^{3} e^{u} \cdot 2 u d u

= \int_{1}^{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_{1}^{3} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{1}^{3}

= 2 [e^{u} u - \int e^{u} d u]_{1}^{3}

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

= 2 [e^{u} u - e^{u}]_{1}^{3}

Compute the boundaries:

2 e^{3}

= 2 \cdot 2 e^{3}

Simplify

= 4 e^{3}

k = 1 \sum \infty \frac{1}{k ^{4}}

4 x^{2} cos (x) + 3 x

\int 4 e^{x} sec (e^{x} + 2) d x

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

\frac{1}{s ^{\frac{3}{2}}}

What is the integral from 1 to 9 of e^{sqrt(x)} ?
The integral from 1 to 9 of e^{sqrt(x)} is 4e^3