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

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

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

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

\frac{c ( - y ^{2} + 1 )}{2}

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{1 - y^{2}}

= c [\frac{x ^{2}}{2}]_{0}^{1 - y^{2}}

Compute the boundaries:

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

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

Simplify

= \frac{c ( - y ^{2} + 1 )}{2}

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