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

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

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

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

- \frac{8}{3}

Decimal

- 2.66666 \dots

Solution steps

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

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{16}{3}

= - \frac{1}{2} \cdot \frac{16}{3}

Simplify

= - \frac{8}{3}

\int_{7}^{\infty} \frac{1}{x ^{2} + x} d x

\int 2 x x^{2} - 4 d x

f (x) = \frac{1}{x}, [1, 3]

\int_{0}^{2} t^{3} (2 + t^{4})^{3} d t

\int_{0}^{8} 2 π y (y^{\frac{2}{3}}) d y

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