Science:Math Exam Resources/Courses/MATH101/April 2012/Question 03 (b)/Solution 1

We can write the integral as $\int _{0}^{3}(x+1){\sqrt {9-x^{2}}}dx=\int _{0}^{3}x{\sqrt {9-x^{2}}}dx+\int _{0}^{3}{\sqrt {9-x^{2}}}dx$.

Now we compute: for $\int _{0}^{3}x{\sqrt {9-x^{2}}}dx$ we use the substitution $u=9-x^{2},$ $du=-2xdx$ to get: $\int _{0}^{3}x{\sqrt {9-x^{2}}}dx=-\int _{9}^{0}{\frac {1}{2}}{\sqrt {u}}du=-{\frac {1}{3}}u^{3/2}|_{9}^{0}=-(0-{\frac {1}{3}}9^{3/2})=9$

For $\int _{0}^{3}{\sqrt {9-x^{2}}}dx$ we note that this corresponds to one fourth of the area of a circle of radius 3, so $\int _{0}^{3}{\sqrt {9-x^{2}}}dx={\frac {1}{4}}9\pi$

Hence the final result is $\int _{0}^{3}(x+1){\sqrt {9-x^{2}}}dx=9+{\frac {9\pi }{4}}$