What is the integral from-2 to 0 of 3xsqrt(4-x^2) ?

The integral from-2 to 0 of 3xsqrt(4-x^2) is -8

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integral from-2 to 0 of 3xsqrt(4-x^2)

\int_{- 2}^{0} 3 x 4 - x^{2} d x

- 8

Solution steps

\int_{- 2}^{0} 3 x 4 - x^{2} d x

Take the constant out:

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

= 3 \cdot \int_{- 2}^{0} x 4 - x^{2} d x

Apply u-substitution

: \int_{0}^{4} - \frac{u}{2} d u

= 3 \cdot \int_{0}^{4} - \frac{u}{2} d u

Take the constant out:

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

= 3 (- \frac{1}{2} \cdot \int_{0}^{4} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= 3 (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4})

Simplify

3 (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}) : - \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

= - \frac{3}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{4}

Compute the boundaries:

\frac{16}{3}

= - \frac{3}{2} \cdot \frac{16}{3}

Simplify

= - 8

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

\frac{\partial}{\partial y} (ln (y - 3 x))

\int - 24 t e^{- 3 t} d t

\int \frac{x ^{2} - x + 2}{x ^{4} + 4 x ^{2} + 3} d x

What is the integral from-2 to 0 of 3xsqrt(4-x^2) ?
The integral from-2 to 0 of 3xsqrt(4-x^2) is -8