What is the integral from-infinity to 6 of re^{r/2} ?

The integral from-infinity to 6 of re^{r/2} is 8e^3

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integral from-infinity to 6 of re^{r/2}

\int_{- \infty}^{6} r e^{\frac{r}{2}} d r

8 e^{3}

Decimal

160.68429 \dots

Solution steps

\int_{- \infty}^{6} r e^{\frac{r}{2}} d r

Compute the indefinite integral:

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

Compute the boundaries:

\int_{- \infty}^{6} r e^{\frac{r}{2}} d r = 8 e^{3} - 0

= 8 e^{3} - 0

Simplify

= 8 e^{3}

\frac{\partial}{\partial x} (4 y^{3})

\frac{d y}{d t} = 3 y + e^{7 t}

7 (5^{x})

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

f (x) = 5 x cos (x), at x = π

What is the integral from-infinity to 6 of re^{r/2} ?
The integral from-infinity to 6 of re^{r/2} is 8e^3