What is the integral from 7 to 9 of |f(4x)| ?

The integral from 7 to 9 of |f(4x)| is -64f

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integral from 7 to 9 of |f(4x)|

\int_{7}^{9} ∣ f (4 x) ∣ d x

- 64 f

Solution steps

\int_{7}^{9} ∣ f \cdot 4 x ∣ d x

Eliminate Absolutes

= \int_{7}^{9} - 4 f x d x

Take the constant out:

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

= - 4 f \cdot \int_{7}^{9} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{7}^{9}

= - 4 f [\frac{x ^{2}}{2}]_{7}^{9}

Compute the boundaries:

16

= - 4 f \cdot 16

Simplify

= - 64 f

\int_{1}^{2} (7 - 6 x)^{2} d x

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

\int_{0}^{0.5} \frac{1}{( 1 - x ) ^{2}} d x

\int_{1}^{1.5} 1 + 36 x d x

\int_{0}^{4} x + 2 - x d x

What is the integral from 7 to 9 of |f(4x)| ?
The integral from 7 to 9 of |f(4x)| is -64f