What is the integral of xsqrt(16-25x^2) ?

The integral of xsqrt(16-25x^2) is -1/75 (16-25x^2)^{3/2}+C

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integral of xsqrt(16-25x^2)

\int x 16 - 25 x^{2} d x

- \frac{1}{75} (16 - 25 x^{2})^{\frac{3}{2}} + C

Solution steps

\int x 16 - 25 x^{2} d x

Apply u-substitution

: \int - \frac{u}{50} d u

= \int - \frac{u}{50} d u

Take the constant out:

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

= - \frac{1}{50} \cdot \int u d u

Apply the Power Rule

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

= - \frac{1}{50} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 16 - 25 x^{2}

= - \frac{1}{50} \cdot \frac{2}{3} (16 - 25 x^{2})^{\frac{3}{2}}

Simplify

- \frac{1}{50} \cdot \frac{2}{3} (16 - 25 x^{2})^{\frac{3}{2}} : - \frac{1}{75} (16 - 25 x^{2})^{\frac{3}{2}}

= - \frac{1}{75} (16 - 25 x^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{75} (16 - 25 x^{2})^{\frac{3}{2}} + C

\int_{0}^{\frac{π}{2}} 3 cos (x) d x

f (x) = (3 x^{3} + 2 x) (x - 2) (x + 4)

\int \frac{1}{( 4 x ^{2} - 25 ) ^{\frac{3}{2}}} d x

\int - 4 x^{2} - 12 x - 5 d x

x \to 0 + lim (\frac{2}{x})

What is the integral of xsqrt(16-25x^2) ?
The integral of xsqrt(16-25x^2) is -1/75 (16-25x^2)^{3/2}+C