What is integral from 0 to 4 of f(x) ?

The answer to integral from 0 to 4 of f(x) is f*8

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integral from 0 to 4 of f(x)

\int_{0}^{4} f (x) d x

f \cdot 8

Solution steps

\int_{0}^{4} f x d x

Take the constant out:

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

= f \cdot \int_{0}^{4} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{4}

= f [\frac{x ^{2}}{2}]_{0}^{4}

Compute the boundaries:

8

= f \cdot 8

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

n = 0 \sum \infty \frac{1}{( 2 n + 1 ) ^{4}}

x + 3 y = 34, x + 6 = y^{2}

\frac{d}{d x} (9 x^{2} e^{- x})

f (x) = \frac{5}{2} + \frac{3}{2} x

What is integral from 0 to 4 of f(x) ?
The answer to integral from 0 to 4 of f(x) is f*8