What is the integral from 1 to 2 of 5e^{1/x} ?

The integral from 1 to 2 of 5e^{1/x} is 5(-e+\Ei(1)+2sqrt(e)-\Ei(1/2))

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integral from 1 to 2 of 5e^{1/x}

\int_{1}^{2} 5 e^{\frac{1}{x}} d x

5 (- e + Ei (1) + 2 e - Ei (\frac{1}{2}))

Solution steps

\int_{1}^{2} 5 e^{\frac{1}{x}} d x

Take the constant out:

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

= 5 \cdot \int_{1}^{2} e^{\frac{1}{x}} d x

Apply u-substitution

: \int_{1}^{\frac{1}{2}} - \frac{e ^{u}}{u ^{2}} d u

= 5 \cdot \int_{1}^{\frac{1}{2}} - \frac{e ^{u}}{u ^{2}} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 5 (- \int_{\frac{1}{2}}^{1} - \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

= 5 (- (- \int_{\frac{1}{2}}^{1} \frac{e ^{u}}{u ^{2}} d u))

Apply Integration By Parts

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

= 5 (- (- [- \frac{e ^{u}}{u} - \int - \frac{e ^{u}}{u} d u]_{\frac{1}{2}}^{1}))

\int - \frac{e ^{u}}{u} d u = - Ei (u)

= 5 (- (- [- \frac{e ^{u}}{u} - (- Ei (u))]_{\frac{1}{2}}^{1}))

Simplify

= 5 [- \frac{e ^{u}}{u} + Ei (u)]_{\frac{1}{2}}^{1}

Compute the boundaries:

- e + Ei (1) + 2 e - Ei (\frac{1}{2})

= 5 (- e + Ei (1) + 2 e - Ei (\frac{1}{2}))

\frac{20}{( s ^{2} + 4 ) ( s - 4 )}

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

x \to 2 lim (tan (\frac{π x}{4}))

\frac{\partial}{\partial x} (ln (\frac{2 x y}{7 z}))

x \to \infty lim (\frac{x}{5})

What is the integral from 1 to 2 of 5e^{1/x} ?
The integral from 1 to 2 of 5e^{1/x} is 5(-e+\Ei(1)+2sqrt(e)-\Ei(1/2))