What is the integral of re^{r/2} ?

The integral of re^{r/2} is 4(1/2 e^{r/2}r-e^{r/2})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of re^{r/2}

\int r e^{\frac{r}{2}} d r

4 (\frac{1}{2} e^{\frac{r}{2}} r - e^{\frac{r}{2}}) + C

Solution steps

\int r e^{\frac{r}{2}} d r

Apply u-substitution

: \int 4 e^{u} u d u

= \int 4 e^{u} u d u

Take the constant out:

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

= 4 \cdot \int e^{u} u d u

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= 4 (e^{u} u - \int e^{u} d u)

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

= 4 (e^{u} u - e^{u})

Substitute back

u = \frac{r}{2}

= 4 (e^{\frac{r}{2}} \frac{r}{2} - e^{\frac{r}{2}})

Simplify

4 (e^{\frac{r}{2}} \frac{r}{2} - e^{\frac{r}{2}}) : 4 (\frac{1}{2} e^{\frac{r}{2}} r - e^{\frac{r}{2}})

= 4 (\frac{1}{2} e^{\frac{r}{2}} r - e^{\frac{r}{2}})

Add a constant to the solution

= 4 (\frac{1}{2} e^{\frac{r}{2}} r - e^{\frac{r}{2}}) + C

\frac{d}{d x} (x^{- \frac{5}{2}})

\frac{\partial}{\partial x} (∣ x + y ∣)

\int_{0}^{1} (6 x^{e} + 5 e^{x}) d x

\frac{\partial}{\partial x} ((- 13 + 142) \cdot \frac{1}{17})

\frac{\partial}{\partial y} (x^{0.5} \cdot y^{0.5})

What is the integral of re^{r/2} ?
The integral of re^{r/2} is 4(1/2 e^{r/2}r-e^{r/2})+C