Note that since (x-3)^2 = x^2-6x+9, what you have is
∫√(25-(x-3)^2) dx
If u = x-3, du = dx, and it's just
∫√(25-u^2) du
which you should know. If not, just use the trig substitution u = 5sinθ
The answer can be seen at
http://www.wolframalpha.com/input/?i=%E2%88%AB%E2%88%9A%2825-u^2%29+du
Or, in terms of the original integral,
http://www.wolframalpha.com/input/?i=%E2%88%AB%E2%88%9A%2816%2B6x-x^2%29+dx
integral: sqrt (16 + 6x − x^2)
dx
3 answers
thanks. i wish i could upload a screen shot of what i typed it but i did it exactly as the site gave and it was counted wrong.
nevermind! i just forgot the constant