What is the integral from 0 to 2 of 2x^2 ?

The integral from 0 to 2 of 2x^2 is 16/3

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integral from 0 to 2 of 2x^2

\int_{0}^{2} 2 x^{2} d x

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

\int_{0}^{2} 2 x^{2} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{2} x^{2} d x

Apply the Power Rule

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

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

Compute the boundaries:

\frac{8}{3}

= 2 \cdot \frac{8}{3}

Simplify

= \frac{16}{3}

\frac{d}{d x} (1 - 2 x^{2})

\frac{d}{d x} (\frac{x}{x ^{3} + 1})

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

y = sin (sin (x)), (π, 0)

\frac{d}{d x} (sin (x^{2} - 4))

What is the integral from 0 to 2 of 2x^2 ?
The integral from 0 to 2 of 2x^2 is 16/3