What is the integral of 3ysqrt(7-3y^2) ?

The integral of 3ysqrt(7-3y^2) is -1/3 (7-3y^2)^{3/2}+C

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integral of 3ysqrt(7-3y^2)

\int 3 y 7 - 3 y^{2} d y

- \frac{1}{3} (7 - 3 y^{2})^{\frac{3}{2}} + C

Solution steps

\int 3 y 7 - 3 y^{2} d y

Take the constant out:

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

= 3 \cdot \int y 7 - 3 y^{2} d y

Apply u-substitution

: \int - \frac{u}{6} d u

= 3 \cdot \int - \frac{u}{6} d u

Take the constant out:

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

= 3 (- \frac{1}{6} \cdot \int u d u)

Apply the Power Rule

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

= 3 (- \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 7 - 3 y^{2}

= 3 (- \frac{1}{6} \cdot \frac{2}{3} (7 - 3 y^{2})^{\frac{3}{2}})

Simplify

3 (- \frac{1}{6} \cdot \frac{2}{3} (7 - 3 y^{2})^{\frac{3}{2}}) : - \frac{1}{3} (7 - 3 y^{2})^{\frac{3}{2}}

= - \frac{1}{3} (7 - 3 y^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{3} (7 - 3 y^{2})^{\frac{3}{2}} + C

\int_{0}^{1} (y^{2} + y^{4}) d y

\frac{d}{d x} (x^{2} 3 x + 1)

\int \frac{x}{y ^{2} + x ^{2}} d x

x \to 1 lim (\frac{x ^{3} - 1}{x - 6})

x^{e}

What is the integral of 3ysqrt(7-3y^2) ?
The integral of 3ysqrt(7-3y^2) is -1/3 (7-3y^2)^{3/2}+C