What is the integral from-2 to-u^2 of 1/2 ?

The integral from-2 to-u^2 of 1/2 is -(u^2)/2+1

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integral from-2 to-u^2 of 1/2

\int_{- 2}^{- u^{2}} \frac{1}{2} d v

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

Solution steps

\int_{- 2}^{- u^{2}} \frac{1}{2} d v

Integral of a constant:

\int a d x = a x

= [\frac{1}{2} v]_{- 2}^{- u^{2}}

Compute the boundaries:

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

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

Simplify

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

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What is the integral from-2 to-u^2 of 1/2 ?
The integral from-2 to-u^2 of 1/2 is -(u^2)/2+1