What is the integral from-l to 0 of (x+l) ?

The integral from-l to 0 of (x+l) is -(l^2)/2+l^2

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integral from-l to 0 of (x+l)

\int_{- l}^{0} (x + l) d x

- \frac{l ^{2}}{2} + l^{2}

Solution steps

\int_{- l}^{0} x + l 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_{- l}^{0} x d x + \int_{- l}^{0} l d x

\int_{- l}^{0} x d x = - \frac{l ^{2}}{2}

\int_{- l}^{0} l d x = l^{2}

= - \frac{l ^{2}}{2} + l^{2}

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What is the integral from-l to 0 of (x+l) ?
The integral from-l to 0 of (x+l) is -(l^2)/2+l^2