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Popular Calculus >

integral of (t-1)

Solution

\int (t - 1) d t

Solution

\frac{t ^{2}}{2} - t + C

Solution steps

\int t - 1 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 t d t - \int 1 d t

\int t d t = \frac{t ^{2}}{2}

\int 1 d t = t

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

Add a constant to the solution

= \frac{t ^{2}}{2} - t + C

Graph

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Frequently Asked Questions (FAQ)

What is the integral of (t-1) ?
The integral of (t-1) is (t^2)/2-t+C