What is the integral of sqrt(25-x^2)(2x) ?

The integral of sqrt(25-x^2)(2x) is -2/3 (25-x^2)^{3/2}+C

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integral of sqrt(25-x^2)(2x)

\int 25 - x^{2} (2 x) d x

- \frac{2}{3} (25 - x^{2})^{\frac{3}{2}} + C

Solution steps

\int 25 - x^{2} \cdot 2 x d x

Take the constant out:

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

= 2 \cdot \int 25 - x^{2} x 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 = 25 - x^{2}

= 2 (- \frac{1}{2} \cdot \frac{2}{3} (25 - x^{2})^{\frac{3}{2}})

Simplify

2 (- \frac{1}{2} \cdot \frac{2}{3} (25 - x^{2})^{\frac{3}{2}}) : - \frac{2}{3} (25 - x^{2})^{\frac{3}{2}}

= - \frac{2}{3} (25 - x^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{3} (25 - x^{2})^{\frac{3}{2}} + C

\frac{\partial}{\partial x} (\frac{- t ^{2}}{( 1 - x ^{2} )})

y^{^{'}} - 2 y = x + 5

x \to \frac{5 π}{2} - lim (sec (x))

\int \frac{y}{z ^{2}} d z

\frac{d}{d x} (x - 1)

What is the integral of sqrt(25-x^2)(2x) ?
The integral of sqrt(25-x^2)(2x) is -2/3 (25-x^2)^{3/2}+C