What is the integral of 2x(sqrt(4-x^2)) ?

The integral of 2x(sqrt(4-x^2)) is -2/3 (4-x^2)^{3/2}+C

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integral of 2x(sqrt(4-x^2))

integral

2 x (4 - x^{2})

- \frac{2}{3} (4 - x^{2})^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int - \frac{u}{2} d u

= 2 \cdot \int - \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 u d u)

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= 2 (- \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 4 - x^{2}

= 2 (- \frac{1}{2} \cdot \frac{2}{3} (4 - x^{2})^{\frac{3}{2}})

Simplify

2 (- \frac{1}{2} \cdot \frac{2}{3} (4 - x^{2})^{\frac{3}{2}}) : - \frac{2}{3} (4 - x^{2})^{\frac{3}{2}}

= - \frac{2}{3} (4 - x^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{3} (4 - x^{2})^{\frac{3}{2}} + C

\frac{1}{( x ^{- 0} 0.5 + x ^{0.5} )}

\frac{e ^{1}}{e ^{x} + 1}

\frac{ln ( 1 + x ^{2} )}{x ^{2}}

\frac{1}{t ^{2}}

\frac{x}{( x ^{2} + 1 ) ^{\frac{1}{3}}}

What is the integral of 2x(sqrt(4-x^2)) ?
The integral of 2x(sqrt(4-x^2)) is -2/3 (4-x^2)^{3/2}+C