What is the integral of (25-0.2x) ?

The integral of (25-0.2x) is 25x-0.1x^2+C

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integral of (25-0.2x)

\int (25 - 0.2 x) d x

25 x - 0.1 x^{2} + C

Solution steps

\int 25 - 0.2 x d x

Apply the Sum Rule:

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

= \int 25 d x - \int 0.2 x d x

\int 25 d x = 25 x

\int 0.2 x d x = 0.1 x^{2}

= 25 x - 0.1 x^{2}

Add a constant to the solution

= 25 x - 0.1 x^{2} + C

f (x) = \frac{1}{9 x}

\frac{\partial}{\partial x} (t^{4} e^{- x})

\int (\frac{2}{x} - 3 x x + 2) d x

y = - x^{3} + 2 x^{2} + 1, at x = - 1

y^{^{''}} - 2 a y^{^{'}} + a^{2} y = 0