What is the integral of (12.5-0.02x) ?

The integral of (12.5-0.02x) is 12.5x-0.01x^2+C

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integral of (12.5-0.02x)

\int (12.5 - 0.02 x) d x

12.5 x - 0.01 x^{2} + C

Solution steps

\int 12.5 - 0.02 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 12.5 d x - \int 0.02 x d x

\int 12.5 d x = 12.5 x

\int 0.02 x d x = 0.01 x^{2}

= 12.5 x - 0.01 x^{2}

Add a constant to the solution

= 12.5 x - 0.01 x^{2} + C

\int (cos (5 x))^{2} d x

n = 1 \sum \infty \frac{4}{n ^{2} + 3 n + 2}

\frac{d}{d x} (2 e^{\frac{s i n ( x )}{3}} + 1)

sin (\frac{π}{2} - x)

x \to - 1 - lim (x^{2})

What is the integral of (12.5-0.02x) ?
The integral of (12.5-0.02x) is 12.5x-0.01x^2+C