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∫ 2x(b−f⁻¹(x)) dx for monotonic f | JEE

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Question
If f(x)f(x) is a monotonic and differentiable function, then ∫f(a)f(b)2x(b−f−1(x)) dx\displaystyle\int_{f(a)}^{f(b)}2x\big(b-f^{-1}(x)\big)\,dx equals
A∫abf2(x) dx\displaystyle\int_{a}^{b}f^{2}(x)\,dx
B∫ab(f2(x)−f2(a)) dx\displaystyle\int_{a}^{b}\big(f^{2}(x)-f^{2}(a)\big)\,dxcorrect
C∫ab(f2(x)−f2(b)) dx\displaystyle\int_{a}^{b}\big(f^{2}(x)-f^{2}(b)\big)\,dx
D∫ab(f2(x)+f2(b)) dx\displaystyle\int_{a}^{b}\big(f^{2}(x)+f^{2}(b)\big)\,dx
Solution
Step 1: Substitute x=f(t)x=f(t), so f−1(x)=tf^{-1}(x)=t and dx=f′(t) dtdx=f'(t)\,dt. The limits map as x=f(a)⇒t=ax=f(a)\Rightarrow t=a and x=f(b)⇒t=bx=f(b)\Rightarrow t=b:
I=∫ab2f(t)(b−t)f′(t) dt=∫ab(b−t) [2f(t)f′(t)] dt.I=\int_{a}^{b}2f(t)\big(b-t\big)f'(t)\,dt=\int_{a}^{b}(b-t)\,\Big[2f(t)f'(t)\Big]\,dt.
 Step 2: Recognise 2f(t)f′(t)=ddt(f2(t))2f(t)f'(t)=\dfrac{d}{dt}\big(f^{2}(t)\big), so the integral is set up for integration by parts:
I=∫ab(b−t) d ⁣(f2(t)).I=\int_{a}^{b}(b-t)\,d\!\big(f^{2}(t)\big).
 Step 3: Integrate by parts with u=(b−t)u=(b-t), dv=d ⁣(f2(t))dv=d\!\big(f^{2}(t)\big):
I=[(b−t)f2(t)]ab−∫abf2(t) (−1) dt=[0−(b−a)f2(a)]+∫abf2(t) dt.I=\Big[(b-t)f^{2}(t)\Big]_{a}^{b}-\int_{a}^{b}f^{2}(t)\,(-1)\,dt=\Big[0-(b-a)f^{2}(a)\Big]+\int_{a}^{b}f^{2}(t)\,dt.
 Step 4: Write (b−a)f2(a)=∫abf2(a) dt(b-a)f^{2}(a)=\displaystyle\int_{a}^{b}f^{2}(a)\,dt to merge the terms:
I=∫abf2(t) dt−∫abf2(a) dt=∫ab(f2(x)−f2(a)) dx.I=\int_{a}^{b}f^{2}(t)\,dt-\int_{a}^{b}f^{2}(a)\,dt=\int_{a}^{b}\big(f^{2}(x)-f^{2}(a)\big)\,dx.
 Correct answer: (2)
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