What is the integral from-1 to 2.5 of f(t) ?

The integral from-1 to 2.5 of f(t) is f*2.625

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integral from-1 to 2.5 of f(t)

\int_{- 1}^{2.5} f (t) d t

f \cdot 2.625

Solution steps

\int_{- 1}^{2.5} 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_{- 1}^{2.5} t d t

Apply the Power Rule

: [\frac{t ^{2}}{2}]_{- 1}^{2.5}

= f [\frac{t ^{2}}{2}]_{- 1}^{2.5}

Simplify

= f [0.5 t^{2}]_{- 1}^{2.5}

Compute the boundaries:

2.625

= f \cdot 2.625

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What is the integral from-1 to 2.5 of f(t) ?
The integral from-1 to 2.5 of f(t) is f*2.625