What is the integral from a to b of 1/2 θ^2 ?

The integral from a to b of 1/2 θ^2 is (b^3-a^3)/6

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integral from a to b of 1/2 θ^2

\int_{a}^{b} \frac{1}{2} θ^{2} d θ

\frac{b ^{3} - a ^{3}}{6}

Solution steps

\int_{a}^{b} \frac{1}{2} θ^{2} d θ

Take the constant out:

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

= \frac{1}{2} \cdot \int_{a}^{b} θ^{2} d θ

Apply the Power Rule

= \frac{1}{2} [\frac{θ ^{3}}{3}]_{a}^{b}

Compute the boundaries:

\frac{b ^{3}}{3} - \frac{a ^{3}}{3}

= \frac{1}{2} (\frac{b ^{3}}{3} - \frac{a ^{3}}{3})

Simplify

= \frac{b ^{3} - a ^{3}}{6}

What is the integral from a to b of 1/2 θ^2 ?
The integral from a to b of 1/2 θ^2 is (b^3-a^3)/6