What is the integral from 0 to 1-cos(θ) of 2r^2 ?

The integral from 0 to 1-cos(θ) of 2r^2 is (2(-cos(θ)+1)^3)/3

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integral from 0 to 1-cos(θ) of 2r^2

\int_{0}^{1 - c o s (θ)} 2 r^{2} d r

\frac{2 ( - cos ( θ ) + 1 ) ^{3}}{3}

Solution steps

\int_{0}^{1 - c o s (θ)} 2 r^{2} d r

Take the constant out:

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

= 2 \cdot \int_{0}^{1 - c o s (θ)} r^{2} d r

Apply the Power Rule

: [\frac{r ^{3}}{3}]_{0}^{1 - c o s (θ)}

= 2 [\frac{r ^{3}}{3}]_{0}^{1 - c o s (θ)}

Compute the boundaries:

\frac{( - cos ( θ ) + 1 ) ^{3}}{3}

= 2 \cdot \frac{( - cos ( θ ) + 1 ) ^{3}}{3}

Simplify

= \frac{2 ( - cos ( θ ) + 1 ) ^{3}}{3}

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What is the integral from 0 to 1-cos(θ) of 2r^2 ?
The integral from 0 to 1-cos(θ) of 2r^2 is (2(-cos(θ)+1)^3)/3