What is the integral of 7.9t-16.6 ?

The integral of 7.9t-16.6 is 3.95t^2-16.6t+C

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integral of 7.9t-16.6

\int 7.9 t - 16.6 d t

3.95 t^{2} - 16.6 t + C

Solution steps

\int 7.9 t - 16.6 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 7.9 t d t - \int 16.6 d t

\int 7.9 t d t = 3.95 t^{2}

\int 16.6 d t = 16.6 t

= 3.95 t^{2} - 16.6 t

Add a constant to the solution

= 3.95 t^{2} - 16.6 t + C

\frac{d}{d x} (\frac{1}{( 1 - 2 x ) ^{\frac{3}{2}}})

\frac{\partial}{\partial x} (x^{\frac{2}{3}} + y^{\frac{2}{3}})

\int \frac{1}{64 + x ^{2}} d x

f (x) = x cos (x)

\frac{d}{d x} (ln (x) + 2 x)

What is the integral of 7.9t-16.6 ?
The integral of 7.9t-16.6 is 3.95t^2-16.6t+C