What is the integral from-infinity to 6 of r*e^{r/3} ?

The integral from-infinity to 6 of r*e^{r/3} is 9e^2

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integral from-infinity to 6 of r*e^{r/3}

\int_{- \infty}^{6} r \cdot e^{\frac{r}{3}} d r

9 e^{2}

Decimal Notation

66.50150 \dots

Solution steps

\int_{- \infty}^{6} 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}^{6} r e^{\frac{r}{3}} d r = 9 e^{2} - 0

= 9 e^{2} - 0

Simplify

= 9 e^{2}

x \to 0 + lim (x ln (2 x))

\frac{d y}{d x}, y = 5 x^{2}

f (x) = \frac{cos ( x ) - 4}{2 sin ( x )}

\frac{1}{3} t^{3} + 7 t^{2} + 50 t

\frac{d}{d x} ((x^{4} + 3 x^{2}) (1 - 5 x))

What is the integral from-infinity to 6 of r*e^{r/3} ?
The integral from-infinity to 6 of r*e^{r/3} is 9e^2