What is the integral from 0 to 3/4 of 60x ?

The integral from 0 to 3/4 of 60x is 135/8

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integral from 0 to 3/4 of 60x

\int_{0}^{\frac{3}{4}} 60 x d x

\frac{135}{8}

Decimal Notation

16.875

Solution steps

\int_{0}^{\frac{3}{4}} 60 x d x

Take the constant out:

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

= 60 \cdot \int_{0}^{\frac{3}{4}} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{\frac{3}{4}}

= 60 [\frac{x ^{2}}{2}]_{0}^{\frac{3}{4}}

Compute the boundaries:

\frac{9}{32}

= 60 \cdot \frac{9}{32}

Simplify

= \frac{135}{8}

y^{^{'}} - 3 y + 2 x = 0

x \to \infty lim (x^{2} - 4)

\int_{0}^{1} \frac{y}{e ^{7 y}} d y

\frac{d}{d x} ((x^{2} + y^{2})^{\frac{4}{5}})

\int e^{3 x} - 1 d x

What is the integral from 0 to 3/4 of 60x ?
The integral from 0 to 3/4 of 60x is 135/8