What is the integral from 0 to 0.5 of 400x ?

The integral from 0 to 0.5 of 400x is 50

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integral from 0 to 0.5 of 400x

\int_{0}^{0.5} 400 x d x

50

Solution steps

\int_{0}^{0.5} 400 x d x

Take the constant out:

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

= 400 \cdot \int_{0}^{0.5} x d x

Apply the Power Rule

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

= 400 [\frac{x ^{2}}{2}]_{0}^{0.5}

Simplify

= 400 [0.5 x^{2}]_{0}^{0.5}

Compute the boundaries:

0.125

= 400 \cdot 0.125

Simplify

= 50

f (x) = (1 + 2 x)^{11}, (0, 1)

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

\frac{\partial}{\partial x} (\frac{x z}{y ^{2} - z})

sec (x)

\int \frac{x}{x ^{3} + 1} d x

What is the integral from 0 to 0.5 of 400x ?
The integral from 0 to 0.5 of 400x is 50