What is the integral from 0 to 16 of f(t) ?

The integral from 0 to 16 of f(t) is f*128

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integral from 0 to 16 of f(t)

\int_{0}^{16} f (t) d t

f \cdot 128

Solution steps

\int_{0}^{16} f t d t

Take the constant out:

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

= f \cdot \int_{0}^{16} t d t

Apply the Power Rule

: [\frac{t ^{2}}{2}]_{0}^{16}

= f [\frac{t ^{2}}{2}]_{0}^{16}

Compute the boundaries:

128

= f \cdot 128

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What is the integral from 0 to 16 of f(t) ?
The integral from 0 to 16 of f(t) is f*128