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

The integral from 0 to 2 of-2xsqrt(4-x^2) is -16/3

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

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

- \frac{16}{3}

Decimal Notation

- 5.33333 \dots

Solution steps

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

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - 2 (- \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

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

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{16}{3}

= - \frac{16}{3}

\int (2 x + 3) ln (x) d x

13 x e^{x}

x \to 0 lim (14 x + 7 h)

\int 2 x arctan (x) d x

\frac{d y}{d x} = \frac{x}{10 y}

What is the integral from 0 to 2 of-2xsqrt(4-x^2) ?
The integral from 0 to 2 of-2xsqrt(4-x^2) is -16/3