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

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

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

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

2 - 3

Decimal Notation

0.26794 \dots

Solution steps

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

Apply u-substitution

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

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

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

= - \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})

Simplify

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

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

Compute the boundaries:

4 - 23

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

Simplify

= 2 - 3

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 5 y = 0

\frac{d}{d x} (\frac{sin ^{2} ( x )}{cos ^{2} ( x )})

\int x^{a} d x

x \to 3 lim (- x^{2} + 5 x - 2)

\frac{d}{d x} (arcsin (x^{2} + y^{2} - 2))

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