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

JEE Maths question with a full step-by-step solution.

Question

f(x)

is a monotonic and differentiable function, then

\displaystyle\int_{f(a)}^{f(b)}2x\big(b-f^{-1}(x)\big)\,dx

equals

\displaystyle\int_{a}^{b}f^{2}(x)\,dx

\displaystyle\int_{a}^{b}\big(f^{2}(x)-f^{2}(a)\big)\,dx

correct

\displaystyle\int_{a}^{b}\big(f^{2}(x)-f^{2}(b)\big)\,dx

\displaystyle\int_{a}^{b}\big(f^{2}(x)+f^{2}(b)\big)\,dx

Solution

Step 1: Substitute

x=f(t)

, so

f^{-1}(x)=t

and

dx=f'(t)\,dt

. The limits map as

x=f(a)\Rightarrow t=a

and

x=f(b)\Rightarrow t=b

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)=\dfrac{d}{dt}\big(f^{2}(t)\big)

, so the integral is set up for integration by parts:

I=\int_{a}^{b}(b-t)\,d\!\big(f^{2}(t)\big).

Step 3: Integrate by parts with

u=(b-t)

dv=d\!\big(f^{2}(t)\big)

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)f^{2}(a)=\displaystyle\int_{a}^{b}f^{2}(a)\,dt

to merge the terms:

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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