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Evaluate the integral by interpreting it in terms of areas. S^0 on the bottom -6 (5+root36-x^2)dx I have no idea how to work this problem out, please help!
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Take a good look at the integrand, does it look familiar to you ?
the -6 is on the bottom of the integral, um no not really
like this ?\[\large \int\limits_{-6}^0 \color{blue}{5+\sqrt{36-x^2}}\, dx\]
yes
May be rearrange the integrand a bit \[\large y=\color{blue}{5+\sqrt{36-x^2}}\] \[\large (y\color{blue}{-5})^2=(\color{blue}{\sqrt{36-x^2}})^2\] \[\large \color{blue}{x^2}+(y\color{blue}{-5})^2=\color{blue}{36}\] what about now ? seen this before ?
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it has to do with circles right?
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