What is the integral from 0 to 0.5 of 600x ?

The integral from 0 to 0.5 of 600x is 75

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integral from 0 to 0.5 of 600x

\int_{0}^{0.5} 600 x d x

75

Solution steps

\int_{0}^{0.5} 600 x d x

Take the constant out:

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

= 600 \cdot \int_{0}^{0.5} x d x

Apply the Power Rule

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

= 600 [\frac{x ^{2}}{2}]_{0}^{0.5}

Simplify

= 600 [0.5 x^{2}]_{0}^{0.5}

Compute the boundaries:

0.125

= 600 \cdot 0.125

Simplify

= 75

f (x) = 5 x^{2} - 6 x + 6 + ln (x)

\frac{d ^{2}}{d x ^{2}} (4 y^{2} + 7 x^{2})

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

\int x sin (x^{2}) d x

\int x^{2} e^{- x^{3}} d x

What is the integral from 0 to 0.5 of 600x ?
The integral from 0 to 0.5 of 600x is 75