What is the integral from 0 to 1.75 of 1250x ?

The integral from 0 to 1.75 of 1250x is 1914.0625

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integral from 0 to 1.75 of 1250x

\int_{0}^{1.75} 1250 x d x

1914.0625

Solution steps

\int_{0}^{1.75} 1250 x d x

Take the constant out:

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

= 1250 \cdot \int_{0}^{1.75} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{1.75}

= 1250 [\frac{x ^{2}}{2}]_{0}^{1.75}

Simplify

= 1250 [0.5 x^{2}]_{0}^{1.75}

Compute the boundaries:

1.53125

= 1250 \cdot 1.53125

Simplify

= 1914.0625

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

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

\frac{d y}{d x} + 4 e^{x + y} = 0

y = x^{2} - 9, y = (2 x - 1) (x + 3)

\frac{d}{d x} (x e^{2 x y})

What is the integral from 0 to 1.75 of 1250x ?
The integral from 0 to 1.75 of 1250x is 1914.0625