What is the integral from 3 to 5 of 3xsqrt(x^2-9) ?

The integral from 3 to 5 of 3xsqrt(x^2-9) is 64

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integral from 3 to 5 of 3xsqrt(x^2-9)

\int_{3}^{5} 3 x x^{2} - 9 d x

64

Solution steps

\int_{3}^{5} 3 x x^{2} - 9 d x

Take the constant out:

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

= 3 \cdot \int_{3}^{5} x x^{2} - 9 d x

Apply u-substitution

: \int_{0}^{4} u^{2} d u

= 3 \cdot \int_{0}^{4} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{0}^{4}

= 3 [\frac{u ^{3}}{3}]_{0}^{4}

Compute the boundaries:

\frac{64}{3}

= 3 \cdot \frac{64}{3}

Simplify

= 64

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What is the integral from 3 to 5 of 3xsqrt(x^2-9) ?
The integral from 3 to 5 of 3xsqrt(x^2-9) is 64