What is the integral of 13-(x/(40)) ?

The integral of 13-(x/(40)) is 13x-(x^2)/(80)+C

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integral of 13-(x/(40))

\int 13 - (\frac{x}{40}) d x

13 x - \frac{x ^{2}}{80} + C

Solution steps

\int 13 - \frac{x}{40} 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 13 d x - \int \frac{x}{40} d x

\int 13 d x = 13 x

\int \frac{x}{40} d x = \frac{x ^{2}}{80}

= 13 x - \frac{x ^{2}}{80}

Add a constant to the solution

= 13 x - \frac{x ^{2}}{80} + C

\int x^{2} ln (6 x) d x

\int 3 x^{2} + 4 x - 1 d x

\frac{d}{d x} (x y)

(4 x + y e^{\frac{y}{x}}) d x - x e^{\frac{y}{x}} d y = 0

What is the integral of 13-(x/(40)) ?
The integral of 13-(x/(40)) is 13x-(x^2)/(80)+C