What is the integral from 0 to 0.6 of 2500x ?

The integral from 0 to 0.6 of 2500x is 450

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integral from 0 to 0.6 of 2500x

\int_{0}^{0.6} 2500 x d x

450

Solution steps

\int_{0}^{0.6} 2500 x d x

Take the constant out:

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

= 2500 \cdot \int_{0}^{0.6} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{0.6}

= 2500 [\frac{x ^{2}}{2}]_{0}^{0.6}

Simplify

= 2500 [0.5 x^{2}]_{0}^{0.6}

Compute the boundaries:

0.18

= 2500 \cdot 0.18

Simplify

= 450

(x^{2} + 4) d y = (2 x - 10 x y) d x

x \to 0 + lim (\frac{( ln ( x ) )}{x})

\frac{d}{d x} (\frac{3}{16 + x ^{4}})

\frac{\partial}{\partial θ} (2 cos (θ))

\int \frac{- 4}{x} d x

What is the integral from 0 to 0.6 of 2500x ?
The integral from 0 to 0.6 of 2500x is 450