What is the integral from 0 to 2-y of 5/8 x^3 ?

The integral from 0 to 2-y of 5/8 x^3 is (5(-y+2)^4)/(32)

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integral from 0 to 2-y of 5/8 x^3

\int_{0}^{2 - y} \frac{5}{8} x^{3} d x

\frac{5 ( - y + 2 ) ^{4}}{32}

Solution steps

\int_{0}^{2 - y} \frac{5}{8} x^{3} d x

Take the constant out:

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

= \frac{5}{8} \cdot \int_{0}^{2 - y} x^{3} d x

Apply the Power Rule

= \frac{5}{8} [\frac{x ^{4}}{4}]_{0}^{2 - y}

Compute the boundaries:

\frac{( - y + 2 ) ^{4}}{4}

= \frac{5}{8} \cdot \frac{( - y + 2 ) ^{4}}{4}

Simplify

= \frac{5 ( - y + 2 ) ^{4}}{32}

\frac{d}{d x} (cos (2 x) sin (\frac{x}{3}))

x \to + 4 lim (\frac{3 x}{4} + 2)

\frac{\partial}{\partial x} (2 e^{x^{8} y})

x \to 0 lim (x cos (\frac{9}{x}))

\frac{d ^{2}}{d x ^{2}} (\frac{7}{8 t ^{5}})

What is the integral from 0 to 2-y of 5/8 x^3 ?
The integral from 0 to 2-y of 5/8 x^3 is (5(-y+2)^4)/(32)