What is the integral of (1/2)^{r^2} ?

The integral of (1/2)^{r^2} is (1/2)^{r^2}θ+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1/2)^{r^2}

\int (\frac{1}{2})^{r^{2}} d θ

(\frac{1}{2})^{r^{2}} θ + C

Solution steps

\int (\frac{1}{2})^{r^{2}} d θ

Integral of a constant:

\int a d x = a x

= (\frac{1}{2})^{r^{2}} θ

Add a constant to the solution

= (\frac{1}{2})^{r^{2}} θ + C

f (x) = \frac{14}{( 1 + e - x ) , at ( 0 , 7 )}

\frac{d}{d x} (ln (\frac{x ^{\frac{4}{3}} cos ( x )}{7 x + 6}))

y = f (x), at x = 1

\int_{0}^{1} \frac{37}{x ^{5}} d x

2 y^{^{''}} + 8 y^{^{'}} + 80 y - 60 cos (8 t) = 0

What is the integral of (1/2)^{r^2} ?
The integral of (1/2)^{r^2} is (1/2)^{r^2}θ+C