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

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

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integral of 4xsqrt(4-x^2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 4 \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

= 4 (- \frac{1}{2} \cdot \int u d u)

Apply the Power Rule

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

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

Substitute back

u = 4 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x y + yz + z x - 3)

y = e^{2 x}, at x = \frac{1}{2} ln (2)

x \to - 1 lim (\frac{∣ x ∣}{x + 1})

\int x arctan (8 x) d x

f (x) = cos^{2} (4 x)

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