What is the integral from 0 to 1/2 of y^2(8y) ?

The integral from 0 to 1/2 of y^2(8y) is 1/8

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integral from 0 to 1/2 of y^2(8y)

\int_{0}^{\frac{1}{2}} y^{2} (8 y) d y

\frac{1}{8}

Decimal Notation

0.125

Solution steps

\int_{0}^{\frac{1}{2}} y^{2} \cdot 8 y d y

Take the constant out:

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

= 8 \cdot \int_{0}^{\frac{1}{2}} y^{2} y d y

Simplify

y^{2} y : y^{3}

= 8 \cdot \int_{0}^{\frac{1}{2}} y^{3} d y

Apply the Power Rule

: [\frac{y ^{4}}{4}]_{0}^{\frac{1}{2}}

= 8 [\frac{y ^{4}}{4}]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{1}{64}

= 8 \cdot \frac{1}{64}

Simplify

= \frac{1}{8}

y = (x^{2} - 1)

f (x) = 5 + 4 x^{2} - 2 x^{3}, at x = a

x \to \infty lim (- x + 2)

\int_{0}^{π} 3 sin^{2} (4 x) d x

\int e^{x} d x

What is the integral from 0 to 1/2 of y^2(8y) ?
The integral from 0 to 1/2 of y^2(8y) is 1/8