What is the integral from 1 to 3 of xsqrt(10x^2-9) ?

The integral from 1 to 3 of xsqrt(10x^2-9) is 364/15

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integral from 1 to 3 of xsqrt(10x^2-9)

\int_{1}^{3} x 10 x^{2} - 9 d x

\frac{364}{15}

Decimal Notation

24.26666 \dots

Solution steps

\int_{1}^{3} x 10 x^{2} - 9 d x

Apply u-substitution

: \int_{1}^{81} \frac{u}{20} d u

= \int_{1}^{81} \frac{u}{20} d u

Take the constant out:

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

= \frac{1}{20} \cdot \int_{1}^{81} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{81}

= \frac{1}{20} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{81}

Compute the boundaries:

\frac{1456}{3}

= \frac{1}{20} \cdot \frac{1456}{3}

Simplify

= \frac{364}{15}

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What is the integral from 1 to 3 of xsqrt(10x^2-9) ?
The integral from 1 to 3 of xsqrt(10x^2-9) is 364/15