What is the integral from-1 to 0 of x/(sqrt(4-x^2)) ?

The integral from-1 to 0 of x/(sqrt(4-x^2)) is -2+sqrt(3)

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integral from-1 to 0 of x/(sqrt(4-x^2))

\int_{- 1}^{0} \frac{x}{4 - x ^{2}} d x

- 2 + 3

Decimal

- 0.26794 \dots

Solution steps

\int_{- 1}^{0} \frac{x}{4 - x ^{2}} d x

Apply u-substitution

: \int_{3}^{4} - \frac{1}{2 u} d u

= \int_{3}^{4} - \frac{1}{2 u} 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_{3}^{4} \frac{1}{u} d u

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{3}^{4}

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

Apply radical rule:

a^{\frac{1}{n}} = n a

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

Compute the boundaries:

4 - 23

= - \frac{1}{2} (4 - 23)

Simplify

= - 2 + 3

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What is the integral from-1 to 0 of x/(sqrt(4-x^2)) ?
The integral from-1 to 0 of x/(sqrt(4-x^2)) is -2+sqrt(3)