What is the integral of 5re^{r/2} ?

The integral of 5re^{r/2} is 20(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 5re^{r/2}

\int 5 r e^{\frac{r}{2}} d r

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

Solution steps

\int 5 r e^{\frac{r}{2}} d r

Take the constant out:

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

= 5 \cdot \int r e^{\frac{r}{2}} d r

Apply u-substitution

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

= 5 \cdot \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

= 5 \cdot 4 \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 5 \cdot 4 (e^{u} u - \int e^{u} d u)

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

= 5 \cdot 4 (e^{u} u - e^{u})

Substitute back

u = \frac{r}{2}

= 5 \cdot 4 (e^{\frac{r}{2}} \frac{r}{2} - e^{\frac{r}{2}})

Simplify

5 \cdot 4 (e^{\frac{r}{2}} \frac{r}{2} - e^{\frac{r}{2}}) : 20 (\frac{1}{2} e^{\frac{r}{2}} r - e^{\frac{r}{2}})

= 20 (\frac{1}{2} e^{\frac{r}{2}} r - e^{\frac{r}{2}})

Add a constant to the solution

= 20 (\frac{1}{2} e^{\frac{r}{2}} r - e^{\frac{r}{2}}) + C

f (x) = x^{2} ln (8 - 5 x^{2})

y = x^{3} - 10 x, y = 6 x

\frac{d y}{d x} = x^{6} y^{- 5}

\int_{- 5}^{5} \frac{1}{∣ x ∣ ^{\frac{2}{3}}} d x

y = arctan (x)

What is the integral of 5re^{r/2} ?
The integral of 5re^{r/2} is 20(1/2 e^{r/2}r-e^{r/2})+C