What is the integral of 10-1/16 x ?

The integral of 10-1/16 x is 10x-(x^2)/(32)+C

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integral of 10-1/16 x

\int 10 - \frac{1}{16} x d x

10 x - \frac{x ^{2}}{32} + C

Solution steps

\int 10 - \frac{1}{16} 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 10 d x - \int \frac{1}{16} x d x

\int 10 d x = 10 x

\int \frac{1}{16} x d x = \frac{x ^{2}}{32}

= 10 x - \frac{x ^{2}}{32}

Add a constant to the solution

= 10 x - \frac{x ^{2}}{32} + C

\int \frac{1}{2 sin ^{2} ( \frac{x}{5} )} d x

x = ln (x y), at x = 1

2 x + 1

\int \frac{2}{cos ^{2} ( x )} d x

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 5, y^{^{'}} (0) = 2

What is the integral of 10-1/16 x ?
The integral of 10-1/16 x is 10x-(x^2)/(32)+C