For every b 1, the area enclosed by the x-axis, the graph of y = f(x), and the vertical lines x = 1 and x = b is given by √(b² + 1) − √2. Then f(x) must be:

Question

For every b > 1, the area enclosed by the x-axis, the graph of y = f(x), and the vertical lines x = 1 and x = b is given by √(b² + 1) − √2. Then f(x) must be:

Accepted Answer

x/√(1 + x²). If the integral of f from 1 to b equals sqrt(b^2+1)-sqrt2, then by the Fundamental Theorem f(b)=d/db[sqrt(b^2+1)]=b/sqrt(b^2+1). Thus f(x)=x/sqrt(1+x^2).

For every b > 1, the area enclosed by the x-axis, the graph of y = f(x), and the vertical lines x = 1 and x = b is given by √(b² + 1) − √2. Then f(x) must be:

Solution

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