What is the integral from 4 to 5 of xsqrt(x^2-16) ?

The integral from 4 to 5 of xsqrt(x^2-16) is 9

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integral from 4 to 5 of xsqrt(x^2-16)

\int_{4}^{5} x x^{2} - 16 d x

9

Solution steps

\int_{4}^{5} x x^{2} - 16 d x

Apply u-substitution

: \int_{0}^{3} u^{2} d u

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

Apply the Power Rule

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

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

Compute the boundaries:

9

= 9

\frac{\partial}{\partial x} (\frac{1}{3} (x^{2} + 2)^{\frac{3}{2}})

x \to \infty lim (9 x^{2} + x - 3)

\int \frac{1}{y ^{3} + y} d y

\int \frac{1}{x ( x ^{2} + 1 ) ^{2}} d x

\int \frac{1}{( x + 1 ) x ^{2} + 2 x - 8} d x

What is the integral from 4 to 5 of xsqrt(x^2-16) ?
The integral from 4 to 5 of xsqrt(x^2-16) is 9