What is the integral from 1 to 6 of f(1/2 x) ?

The integral from 1 to 6 of f(1/2 x) is (35f)/4

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integral from 1 to 6 of f(1/2 x)

\int_{1}^{6} f (\frac{1}{2} x) d x

\frac{35 f}{4}

Solution steps

\int_{1}^{6} f \frac{1}{2} x d x

Take the constant out:

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

= \frac{1}{2} f \cdot \int_{1}^{6} x d x

Simplify

\frac{1}{2} f : \frac{f}{2}

= \frac{f}{2} \cdot \int_{1}^{6} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{1}^{6}

= \frac{f}{2} [\frac{x ^{2}}{2}]_{1}^{6}

Compute the boundaries:

\frac{35}{2}

= \frac{f}{2} \cdot \frac{35}{2}

Simplify

= \frac{35 f}{4}

\frac{d}{d x} (\frac{( 3 x ^{2} - 2 x - 3 )}{( x ^{2} + 1 ) ^{2}})

y^{^{'}} = x - x y

\int e^{s i n (t)} d t

(1.1) (4.2)

\frac{2}{[ ( 4 s ^{2} - 8 s ) ]}

What is the integral from 1 to 6 of f(1/2 x) ?
The integral from 1 to 6 of f(1/2 x) is (35f)/4