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

The integral from 2 to 3 of 2xsqrt(x^2-1) is 2((16sqrt(2))/3-sqrt(3))

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

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

2 (\frac{16 2}{3} - 3)

Decimal

11.62084 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 2 \cdot \int_{3}^{22} u^{2} d u

Apply the Power Rule

= 2 [\frac{u ^{3}}{3}]_{3}^{22}

Compute the boundaries:

\frac{16 2}{3} - 3

= 2 (\frac{16 2}{3} - 3)

What is the integral from 2 to 3 of 2xsqrt(x^2-1) ?
The integral from 2 to 3 of 2xsqrt(x^2-1) is 2((16sqrt(2))/3-sqrt(3))