What is the integral from 2 to 3 of 3xsqrt(2x^2-1) ?

The integral from 2 to 3 of 3xsqrt(2x^2-1) is (17sqrt(17)-7sqrt(7))/2

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integral from 2 to 3 of 3xsqrt(2x^2-1)

\int_{2}^{3} 3 x 2 x^{2} - 1 d x

\frac{17 17 - 7 7}{2}

Decimal Notation

25.78626 \dots

Solution steps

\int_{2}^{3} 3 x 2 x^{2} - 1 d x

Take the constant out:

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

= 3 \cdot \int_{2}^{3} x 2 x^{2} - 1 d x

Apply u-substitution

: \int_{7}^{17} \frac{u}{4} d u

= 3 \cdot \int_{7}^{17} \frac{u}{4} d u

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int_{7}^{17} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{17}

= 3 \cdot \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{17}

Simplify

3 \cdot \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{17} : \frac{3}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{17}

= \frac{3}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{17}

Compute the boundaries:

\frac{34 17 - 14 7}{3}

= \frac{3}{4} \cdot \frac{34 17 - 14 7}{3}

Simplify

= \frac{17 17 - 7 7}{2}

x \to 0 lim ((2 x + 1) (x - 3))

\frac{8}{x}

\frac{\partial}{\partial x} (vx)

y^{^{'}} - 2 y = - 3 e^{t}

x \to - 3 lim (\frac{8}{x + 3})

What is the integral from 2 to 3 of 3xsqrt(2x^2-1) ?
The integral from 2 to 3 of 3xsqrt(2x^2-1) is (17sqrt(17)-7sqrt(7))/2