What is the integral of 5x*sqrt(16-x^2) ?

The integral of 5x*sqrt(16-x^2) is -5/3 (16-x^2)^{3/2}+C

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integral of 5x*sqrt(16-x^2)

\int 5 x \cdot 16 - x^{2} d x

- \frac{5}{3} (16 - x^{2})^{\frac{3}{2}} + C

Solution steps

\int 5 x 16 - x^{2} d x

Take the constant out:

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

= 5 \cdot \int x 16 - x^{2} d x

Apply u-substitution

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

= 5 \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

= 5 (- \frac{1}{2} \cdot \int u d u)

Apply the Power Rule

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

= 5 (- \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 16 - x^{2}

= 5 (- \frac{1}{2} \cdot \frac{2}{3} (16 - x^{2})^{\frac{3}{2}})

Simplify

5 (- \frac{1}{2} \cdot \frac{2}{3} (16 - x^{2})^{\frac{3}{2}}) : - \frac{5}{3} (16 - x^{2})^{\frac{3}{2}}

= - \frac{5}{3} (16 - x^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{5}{3} (16 - x^{2})^{\frac{3}{2}} + C

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

x \to 3 - lim (\frac{ln ( x )}{x - 3})

\frac{\partial}{\partial x} (- sin (y) + e^{x})

\frac{5 x}{x ^{2} + 16}

\frac{d}{d x} (x \cdot tan (x))

What is the integral of 5x*sqrt(16-x^2) ?
The integral of 5x*sqrt(16-x^2) is -5/3 (16-x^2)^{3/2}+C