What is the integral from-2 to 0 of (x^2)/(16/3) ?

The integral from-2 to 0 of (x^2)/(16/3) is 1/2

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integral from-2 to 0 of (x^2)/(16/3)

\int_{- 2}^{0} \frac{x ^{2}}{\frac{16}{3}} d x

\frac{1}{2}

Decimal Notation

0.5

Solution steps

\int_{- 2}^{0} \frac{x ^{2}}{\frac{16}{3}} d x

Take the constant out:

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

= \frac{1}{\frac{16}{3}} \cdot \int_{- 2}^{0} x^{2} d x

Simplify

= \frac{3}{16} \cdot \int_{- 2}^{0} x^{2} d x

Apply the Power Rule

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

= \frac{3}{16} [\frac{x ^{3}}{3}]_{- 2}^{0}

Compute the boundaries:

\frac{8}{3}

= \frac{3}{16} \cdot \frac{8}{3}

Simplify

= \frac{1}{2}

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

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D (X^{9})

7 y y^{^{'}} cos^{2} (x) = sin (x) (3 y^{2} - π)

\frac{\partial}{\partial x} (3 x^{4} y^{2} + 3)

What is the integral from-2 to 0 of (x^2)/(16/3) ?
The integral from-2 to 0 of (x^2)/(16/3) is 1/2