What is the integral from t-2 to 0 of (x+2) ?

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

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integral from t-2 to 0 of (x+2)

\int_{t - 2}^{0} (x + 2) d x

- \frac{( t - 2 ) ^{2}}{2} - 2 (t - 2)

Solution steps

\int_{t - 2}^{0} x + 2 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_{t - 2}^{0} x d x + \int_{t - 2}^{0} 2 d x

\int_{t - 2}^{0} x d x = - \frac{( t - 2 ) ^{2}}{2}

\int_{t - 2}^{0} 2 d x = - 2 (t - 2)

= - \frac{( t - 2 ) ^{2}}{2} - 2 (t - 2)

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