What is the integral from 0 to sqrt(1-y^2 of)(x+y) ?

The integral from 0 to sqrt(1-y^2 of)(x+y) is ysqrt(-y^2+1)+(-y^2+1)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to sqrt(1-y^2 of)(x+y)

\int_{0}^{1 - y^{2}} (x + y) d x

y - y^{2} + 1 + \frac{- y ^{2} + 1}{2}

Solution steps

\int_{0}^{1 - y^{2}} x + y d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1 - y^{2}} y d x + \int_{0}^{1 - y^{2}} x d x

\int_{0}^{1 - y^{2}} y d x = y - y^{2} + 1

\int_{0}^{1 - y^{2}} x d x = \frac{- y ^{2} + 1}{2}

= y - y^{2} + 1 + \frac{- y ^{2} + 1}{2}

\int_{1}^{2} \frac{e ^{x}}{1 + 2 e ^{x}} d x

\int_{3}^{4} 7 x^{3} d x

\int_{\frac{1}{2}}^{1} ln (x) d x

\int_{0}^{\frac{π}{2}} \frac{1}{2} sin (2 θ) d θ

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

What is the integral from 0 to sqrt(1-y^2 of)(x+y) ?
The integral from 0 to sqrt(1-y^2 of)(x+y) is ysqrt(-y^2+1)+(-y^2+1)/2