## Calculus: Early Transcendentals (2nd Edition)

$4-\sqrt 7\approx1.35425$
The problem can be represented by equation $\int\limits_4^5\frac{x}{\sqrt {x^2-9}}*dx$ which we can use u-substitution and the substitution rule for definite integrals on. $u=x^2-9$ $dx=\frac{du}{2x}$ lower limit 4 to $(4)^2-9=7$ upper limit 5 to $(5)^2-9=16$ Now rewrite the equation with these values... $\int\limits_7^{16}\frac{x}{\sqrt u}*\frac{du}{2x}=\int\limits_7^{16}\frac{1}{2\sqrt u}*du=\int\limits_7^{16}\frac{1}{2}u^{-\frac{1}{2}}*du$ Before our next step we'll have to find the antiderivative of $\frac{1}{2}u^{-\frac{1}{2}}$, which is $u^{\frac{1}{2}}+c$ We can now use the fundamental theorem of calculus. $u^{\frac{1}{2}}|_7^16$ $16^{\frac{1}{2}}-7{\frac{1}{2}}=4-\sqrt 7=1.35425\dots$