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If $y = f (x)$ be an invertible function with inverse $g$ and $h (x) = x f (x),$ then $\int_{f (a)}^{f (b)} g (x) d x + \int_{a}^{b} f (x) d x$ is equal to

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h(a)−h(b)

h(b)−h(a)

h(a)+h(b)

bh(b)−ah(a)

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detailed solution

Correct option is B

Put g(x)=t⇒x=f(t)⇒dx=f′(t)dtI=∫ab tf′(t)dt+∫ab f(x)dx=[tf(t)]ab−∫ab 1⋅f(t)dt+∫ab f(x)dx=h(t)]ab=h(b)−h(a)

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If y=f(x) be an invertible function with inverse g and h(x)=xf(x), then ∫f(a)f(b) g(x)dx+∫ab f(x)dx is equal to

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If $y = f (x)$ be an invertible function with inverse $g$ and $h (x) = x f (x),$ then $\int_{f (a)}^{f (b)} g (x) d x + \int_{a}^{b} f (x) d x$ is equal to