What is the integral from 0 to 0.5 of f(2t) ?

The integral from 0 to 0.5 of f(2t) is 0.25f

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integral from 0 to 0.5 of f(2t)

\int_{0}^{0.5} f (2 t) d t

0.25 f

Solution steps

\int_{0}^{0.5} f \cdot 2 t d t

Take the constant out:

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

= 2 f \cdot \int_{0}^{0.5} t d t

Apply the Power Rule

: [\frac{t ^{2}}{2}]_{0}^{0.5}

= 2 f [\frac{t ^{2}}{2}]_{0}^{0.5}

Simplify

= 2 f [0.5 t^{2}]_{0}^{0.5}

Compute the boundaries:

0.125

= 2 f \cdot 0.125

Simplify

= 0.25 f

x \to - 1 lim (x^{2} - x + 1)

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

\int 7 tan^{3} (θ) d θ

e^{x} y \frac{d y}{d x} = e^{- y} + e^{- 3 x - y}

y^{3} - (18 x + 2) + 3 x y^{2} y^{^{'}} = 0

What is the integral from 0 to 0.5 of f(2t) ?
The integral from 0 to 0.5 of f(2t) is 0.25f