What is the integral from 0 to 1 of pi(2x^2)^2 ?

The integral from 0 to 1 of pi(2x^2)^2 is (4pi)/5

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integral from 0 to 1 of pi(2x^2)^2

\int_{0}^{1} π (2 x^{2})^{2} d x

\frac{4 π}{5}

Decimal

2.51327 \dots

Solution steps

\int_{0}^{1} π (2 x^{2})^{2} d x

Take the constant out:

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

= π \cdot \int_{0}^{1} (2 x^{2})^{2} d x

Simplify

(2 x^{2})^{2} : 4 x^{4}

= π \cdot \int_{0}^{1} 4 x^{4} d x

Take the constant out:

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

= π 4 \cdot \int_{0}^{1} x^{4} d x

Apply the Power Rule

: [\frac{x ^{5}}{5}]_{0}^{1}

= π 4 [\frac{x ^{5}}{5}]_{0}^{1}

Compute the boundaries:

\frac{1}{5}

= π 4 \cdot \frac{1}{5}

Simplify

= \frac{4 π}{5}

y^{^{''}} + 4 y = 2 sec (2 t)

4 x^{2} + 5 x, at x = 3

\frac{\partial}{\partial x} (e^{x y} + x + y)

\int tan^{5} (x) d x

\frac{d x}{d y} = \frac{x ^{2} y ^{2}}{1 + x}

What is the integral from 0 to 1 of pi(2x^2)^2 ?
The integral from 0 to 1 of pi(2x^2)^2 is (4pi)/5