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

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

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

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

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{64}{3}

= \frac{64}{3}

x \to \infty lim (h (x))

\int 6 x (3 x^{2} + 5)^{3} d x

\int \frac{1}{- x ^{2} - 2 x} d x

\int \frac{1}{4 x lo g _{3} ( x )} d x

\frac{\partial}{\partial x} (x tan (y))

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