Let f(x) = ∫√(2 – t^2) dt for t ∈ [1,x]. Find the real roots of the equation x^2 – f'(x) = 0.
Let f(x) = ∫√(2 – t2) dt for t ∈ [1x]. Find the real roots of the equation x2 – f'(x) = 0.
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The equation we need to solve is: x^2 – f'(x) = 0. Taking the derivative on both sides, we get: 2x = f'(x). Now, using the fact that f'(x) = √(2 – t2), we can solve the equation for x: x = √(2 – t2). Therefore, the real roots of the equation are √2 and -√2.
