What is the integral from-infinity to 16 of re^{r/4} ?

The integral from-infinity to 16 of re^{r/4} is 48e^4

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integral from-infinity to 16 of re^{r/4}

\int_{- \infty}^{16} r e^{\frac{r}{4}} d r

48 e^{4}

Decimal Notation

2620.71120 \dots

Solution steps

\int_{- \infty}^{16} r e^{\frac{r}{4}} d r

Compute the indefinite integral:

\int r e^{\frac{r}{4}} d r = 4 e^{\frac{r}{4}} r - 16 e^{\frac{r}{4}} + C

Compute the boundaries:

\int_{- \infty}^{16} r e^{\frac{r}{4}} d r = 48 e^{4} - 0

= 48 e^{4} - 0

Simplify

= 48 e^{4}

\int (1 - x^{2})^{- \frac{1}{2}} d x

\frac{d}{d x} (100 x^{\frac{3}{2}})

n = 1 \sum \infty (\frac{4}{9})^{2 n + 1}

f (x) = x^{2} + x, at x = 2

y^{^{''}} + 3 y^{^{'}} + 3 y = 0

What is the integral from-infinity to 16 of re^{r/4} ?
The integral from-infinity to 16 of re^{r/4} is 48e^4