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

= 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

= 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

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