What is the integral from 6 to x of 1/20+x/(40) ?

The integral from 6 to x of 1/20+x/(40) is x/(20)-3/10+1/40 ((x^2)/2-18)

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integral from 6 to x of 1/20+x/(40)

\int_{6}^{x} \frac{1}{20} + \frac{x}{40} d x

\frac{x}{20} - \frac{3}{10} + \frac{1}{40} (\frac{x ^{2}}{2} - 18)

Solution steps

\int_{6}^{x} \frac{1}{20} + \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_{6}^{x} \frac{1}{20} d x + \int_{6}^{x} \frac{x}{40} d x

\int_{6}^{x} \frac{1}{20} d x = \frac{x}{20} - \frac{3}{10}

\int_{6}^{x} \frac{x}{40} d x = \frac{1}{40} (\frac{x ^{2}}{2} - 18)

= \frac{x}{20} - \frac{3}{10} + \frac{1}{40} (\frac{x ^{2}}{2} - 18)

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What is the integral from 6 to x of 1/20+x/(40) ?
The integral from 6 to x of 1/20+x/(40) is x/(20)-3/10+1/40 ((x^2)/2-18)