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Popular Calculus >

integral of 1-t

Solution

\int 1 - t d t

Solution

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

Solution steps

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

\int 1 d t = t

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

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

Add a constant to the solution

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

Graph

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Frequently Asked Questions (FAQ)

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