What is the integral from 0 to 5 of f(t) ?

The integral from 0 to 5 of f(t) is (25f)/2

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integral from 0 to 5 of f(t)

\int_{0}^{5} f (t) d t

\frac{25 f}{2}

Solution steps

\int_{0}^{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_{0}^{5} t d t

Apply the Power Rule

: [\frac{t ^{2}}{2}]_{0}^{5}

= f [\frac{t ^{2}}{2}]_{0}^{5}

Compute the boundaries:

\frac{25}{2}

= f \frac{25}{2}

Simplify

= \frac{25 f}{2}

- π sin (π x)

\frac{d}{d x} ((x^{4} + 1) sech (2 ln (x)))

x \to 6 + lim (f (x))

y^{^{'}} = (x y^{^{'}} + y)

\frac{d}{d x} (\frac{7 x ( 2 x - 20 ) ^{3}}{4 x + 9})

What is the integral from 0 to 5 of f(t) ?
The integral from 0 to 5 of f(t) is (25f)/2