What is the integral from 0 to 1.75 of 600x ?

The integral from 0 to 1.75 of 600x is 918.75

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integral from 0 to 1.75 of 600x

\int_{0}^{1.75} 600 x d x

918.75

Solution steps

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

Apply the Power Rule

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

= 600 [\frac{x ^{2}}{2}]_{0}^{1.75}

Simplify

= 600 [0.5 x^{2}]_{0}^{1.75}

Compute the boundaries:

1.53125

= 600 \cdot 1.53125

Simplify

= 918.75

x \to 0 lim (e^{\frac{8}{x}})

f (y) = \frac{A}{y ^{9}} + B e^{y}

x \to 2 - lim (\frac{3}{x - 2})

\frac{d}{d x} ((cos (x))^{2} + sin (x^{2}))

\int \frac{x}{( x ^{2} + R ^{2} ) ^{\frac{3}{2}}} d x

What is the integral from 0 to 1.75 of 600x ?
The integral from 0 to 1.75 of 600x is 918.75