What is integral from 1 to 9 of f(x) ?

The answer to integral from 1 to 9 of f(x) is f*40

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integral from 1 to 9 of f(x)

\int_{1}^{9} f (x) d x

f \cdot 40

Solution steps

\int_{1}^{9} f x d x

Take the constant out:

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

= f \cdot \int_{1}^{9} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{1}^{9}

= f [\frac{x ^{2}}{2}]_{1}^{9}

Compute the boundaries:

40

= f \cdot 40

x \to - 1 lim (arctan (\frac{x - 1}{x + 1}))

6 x^{3}

(5 - 7 x)^{3}

12 (x - \frac{1}{x})^{2}, at x = 2

\frac{d}{d x} (\frac{1 - x ^{2}}{1 + x ^{2}})

What is integral from 1 to 9 of f(x) ?
The answer to integral from 1 to 9 of f(x) is f*40