What is the integral from 10 to x of (-1)/(75)(x-20) ?

The integral from 10 to x of (-1)/(75)(x-20) is -1/150 (x^2-40x+300)

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integral from 10 to x of (-1)/(75)(x-20)

\int_{10}^{x} \frac{- 1}{75} (x - 20) d x

- \frac{1}{150} (x^{2} - 40 x + 300)

Solution steps

\int_{10}^{x} \frac{- 1}{75} (x - 20) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{- 1}{75} \cdot \int_{10}^{x} x - 20 d x

Simplify

= (- \frac{1}{75}) \cdot \int_{10}^{x} x - 20 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

= (- \frac{1}{75}) (\int_{10}^{x} x d x - \int_{10}^{x} 20 d x)

\int_{10}^{x} x d x = \frac{x ^{2}}{2} - 50

\int_{10}^{x} 20 d x = 20 x - 200

= (- \frac{1}{75}) (\frac{x ^{2}}{2} - 50 - (20 x - 200))

Simplify

(- \frac{1}{75}) (\frac{x ^{2}}{2} - 50 - (20 x - 200)) : - \frac{1}{150} (x^{2} - 40 x + 300)

= - \frac{1}{150} (x^{2} - 40 x + 300)

\int_{- 2}^{2} x^{2} - x d x

\int_{1}^{2} x - 2 d x

\int_{0}^{1} (e^{- x})^{2} d x

\int_{0}^{1} (3 x^{2} - 1) d x

\int_{- 1}^{\infty} 3 e^{- 4 x} d x

What is the integral from 10 to x of (-1)/(75)(x-20) ?
The integral from 10 to x of (-1)/(75)(x-20) is -1/150 (x^2-40x+300)