Suppose f, g are continuous and differentiable on [0,b], are non-negative on [0,b] and f is non constant with f(0) = 0, then the minimum value of ∫a0g(x)f′(x)dx+∫b0g′(x)f(x)dxonaϵ(0,b] is
A
f(a).g(a)
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B
f(b).g(b)
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C
f(a).g(b)
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D
f(b).g(a)
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Solution
The correct option is C f(a).g(b) Integrating by parts the integral ∫a0g(x)f′(x)dx+∫b0g′(x)f(x)dx=g(x)f(x)|a0−∫a0g′(x)f(x)dx+∫b0g′(x)f(x)dx=f(a).g(a)+∫bag′(x)f(x)dx≥f(a).g(a)+∫bag′(x)f(x)dx=f(a).g(b)