What is the integral from-4 to 4 of y/(sqrt(16-y^2)) ?

The integral from-4 to 4 of y/(sqrt(16-y^2)) is 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-4 to 4 of y/(sqrt(16-y^2))

\int_{- 4}^{4} \frac{y}{16 - y ^{2}} d y

0

Solution steps

\int_{- 4}^{4} \frac{y}{16 - y ^{2}} d y

If f(x) is an odd function and continuous at:

[- a, a]

then

\int_{- a}^{a} f (x) d x = 0

Parity of

\frac{y}{16 - y ^{2}} :

Odd

= 0

\int_{- 1}^{2} (3 x^{2} - 2) d x

\int_{0}^{3} y e^{- x y} d y

\int_{0}^{5} \frac{2 x ^{2}}{25} d x

\int_{0}^{1} \frac{1}{( 11 x + 1 ) ^{3}} d x

\int_{- 2}^{1} - x^{2} + 4 x + 5 d x

What is the integral from-4 to 4 of y/(sqrt(16-y^2)) ?
The integral from-4 to 4 of y/(sqrt(16-y^2)) is 0