What is the integral from 1 to 3 of f(2x) ?

The integral from 1 to 3 of f(2x) is 8f

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integral from 1 to 3 of f(2x)

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

8 f

Solution steps

\int_{1}^{3} f \cdot 2 x d x

Take the constant out:

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

= 2 f \cdot \int_{1}^{3} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{1}^{3}

= 2 f [\frac{x ^{2}}{2}]_{1}^{3}

Compute the boundaries:

4

= 2 f \cdot 4

Simplify

= 8 f

t \to 0 lim (ln (t + 1))

y^{^{''}} + 6 y^{^{'}} + 8 y = cos (3 t)

f (x) = 5 x^{3} + 4 x, at x = - 1

\frac{d x}{d t} = 1 + t - x - t x

\int 2 (2 x + 4)^{5} d x

What is the integral from 1 to 3 of f(2x) ?
The integral from 1 to 3 of f(2x) is 8f