What is the integral from 1 to y of (2-t) ?

The integral from 1 to y of (2-t) is 2y-2-(y^2-1)/2

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integral from 1 to y of (2-t)

\int_{1}^{y} (2 - t) d t

2 y - 2 - \frac{y ^{2} - 1}{2}

Solution steps

\int_{1}^{y} 2 - t d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{y} 2 d t - \int_{1}^{y} t d t

\int_{1}^{y} 2 d t = 2 y - 2

\int_{1}^{y} t d t = \frac{y ^{2} - 1}{2}

= 2 y - 2 - \frac{y ^{2} - 1}{2}

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