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

The integral from-2 to 0 of 3wsqrt(4-w^2) is -8

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integral from-2 to 0 of 3wsqrt(4-w^2)

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

- 8

Solution steps

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

Take the constant out:

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

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

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

t \to 0 lim (\frac{e ^{t} - 1}{t})

f (x) = (\frac{x}{7} + \frac{7}{x})^{7}

\frac{d ^{4}}{d x ^{4}} (ln (1 - x))

\int_{1}^{2} \frac{5}{8 x ^{6}} d x

n = 2 \sum \infty \frac{1}{n - 1}

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