What is the integral of 3wsqrt(4-w^2) ?

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

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integral of 3wsqrt(4-w^2)

\int 3 w 4 - w^{2} d w

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

Solution steps

\int 3 w 4 - w^{2} d w

Take the constant out:

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

= 3 \cdot \int w 4 - w^{2} d w

Apply u-substitution

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

= 3 \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

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

Apply the Power Rule

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

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

Substitute back

u = 4 - w^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (e^{y} + x + x^{4} + 5)

5 + 2 x, \frac{( 15 + x )}{3}

\frac{\partial}{\partial x} (2 x^{2} y + 4 y^{3})

y = (x + \frac{1}{x}) (x - \frac{1}{x} + 1)

\frac{\partial}{\partial x} (x^{2} + \frac{x}{y})

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