What is the integral of (4y+2t-5) ?

The integral of (4y+2t-5) is -5t+t^2+4yt+C

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integral of (4y+2t-5)

\int (4 y + 2 t - 5) d t

- 5 t + t^{2} + 4 y t + C

Solution steps

\int 4 y + 2 t - 5 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 5 d t + \int 2 t d t + \int 4 y d t

\int 5 d t = 5 t

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

\int 4 y d t = 4 y t

= - 5 t + t^{2} + 4 y t

Add a constant to the solution

= - 5 t + t^{2} + 4 y t + C