What is the integral from-1 to 1 of 2x^2 ?

The integral from-1 to 1 of 2x^2 is 4/3

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integral from-1 to 1 of 2x^2

\int_{- 1}^{1} 2 x^{2} d x

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

\int_{- 1}^{1} 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_{- 1}^{1} x^{2} d x

Apply the Power Rule

: [\frac{x ^{3}}{3}]_{- 1}^{1}

= 2 [\frac{x ^{3}}{3}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{3}

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

Simplify

= \frac{4}{3}

\frac{d}{d x} (\frac{- 20}{( 2 + x ) ^{2}})

x^{2} - 11, 10 x

(x, y) \to (0, 0) lim (\frac{e ^{x y} - 1}{x y})

\frac{\partial}{\partial x} (x^{4} + 3 y^{3})

x \to 1 lim (x^{3} - 3 x + 2)

What is the integral from-1 to 1 of 2x^2 ?
The integral from-1 to 1 of 2x^2 is 4/3