What is the integral from-infinity to 12 of re^{r/3} ?

The integral from-infinity to 12 of re^{r/3} is 27e^4

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integral from-infinity to 12 of re^{r/3}

\int_{- \infty}^{12} r e^{\frac{r}{3}} d r

27 e^{4}

Decimal Notation

1474.15005 \dots

Solution steps

\int_{- \infty}^{12} r e^{\frac{r}{3}} d r

Compute the indefinite integral:

\int r e^{\frac{r}{3}} d r = 3 e^{\frac{r}{3}} r - 9 e^{\frac{r}{3}} + C

Compute the boundaries:

\int_{- \infty}^{12} r e^{\frac{r}{3}} d r = 27 e^{4} - 0

= 27 e^{4} - 0

Simplify

= 27 e^{4}

\frac{\partial}{\partial t} (x y z^{2} tan (y t))

x^{2} + x y - y^{2} = - 9, (3, 6)

\frac{2 x}{1 + x ^{2}}

F (x) = ln (2 x), y = 0, y = 2

5 sin (x), 5 cos (2 x), 0, \frac{π}{2}

What is the integral from-infinity to 12 of re^{r/3} ?
The integral from-infinity to 12 of re^{r/3} is 27e^4