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

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

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

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

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

Apply the Power Rule

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

Simplify

\frac{1}{4} (- (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{0}^{4})) : \frac{1}{8} [2 u]_{0}^{4}

= \frac{1}{8} [2 u]_{0}^{4}

Compute the boundaries:

4

= \frac{1}{8} \cdot 4

Simplify

= \frac{1}{2}

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