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Calculus and Beyond Homework Help
Solving Improper Integrals: 1/sqrt(9-x^2) 0 to 3
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[QUOTE="scurty, post: 4671307, member: 400421"] That's because you have been doing "proper" integrals up to this point. Improper integrals involve integrating across a point where the function is not defined. In this case the the function is not defined at x = __. The normal procedure is to introduce a variable for that number and take the limit as a approaches that number. In this case, arcsin is defined on [0,1]. But the original function is not defined on [0,3]. [/QUOTE]
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Calculus and Beyond Homework Help
Solving Improper Integrals: 1/sqrt(9-x^2) 0 to 3
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