Q100 / 155 · WBJEE 2026 · Mathematics
Let $f(x)$ be a real valued function which is monotonic and differentiable. Then for any reals $a$ and $b$, $\displaystyle\int_{f(a)}^{f(b)} 2x\{b - f^{-1}(x)\}\,dx =$
- Correct answerA$\displaystyle\int_{a}^{b}(f^2(x) - f^2(a))\,dx$
- B$\displaystyle\int_{a}^{b}(f(x) - f(a))^2\,dx$
- C$\displaystyle\int_{a}^{b}(bf^2(x) - af^2(a))\,dx$
- D$bf^2(b) + f^{-1}(a)$
Solution
Let $I = \displaystyle\int_{f(a)}^{f(b)} 2x\{b - f^{-1}(x)\}\,dx$
Substitute $x = f(t)$, which gives $dx = f'(t)\,dt$.
When $x = f(a)$, $t = a$.
When $x = f(b)$, $t = b$.
The integral becomes:
$I = \displaystyle\int_{a}^{b} 2f(t)\{b - t\}f'(t)\,dt$
Using integration by parts, taking $(b - t)$ as the first function and $2f(t)f'(t)$ as the second function:
$I = \left[ (b - t)f^2(t) \right]_{a}^{b} - \displaystyle\int_{a}^{b} (-1)f^2(t)\,dt$
$I = (0 - (b - a)f^2(a)) + \displaystyle\int_{a}^{b} f^2(t)\,dt$
$I = \displaystyle\int_{a}^{b} f^2(t)\,dt - \displaystyle\int_{a}^{b} f^2(a)\,dt$
$I = \displaystyle\int_{a}^{b} (f^2(x) - f^2(a))\,dx$
Answer: $\displaystyle\int_{a}^{b}(f^2(x) - f^2(a))\,dx$
Substitute $x = f(t)$, which gives $dx = f'(t)\,dt$.
When $x = f(a)$, $t = a$.
When $x = f(b)$, $t = b$.
The integral becomes:
$I = \displaystyle\int_{a}^{b} 2f(t)\{b - t\}f'(t)\,dt$
Using integration by parts, taking $(b - t)$ as the first function and $2f(t)f'(t)$ as the second function:
$I = \left[ (b - t)f^2(t) \right]_{a}^{b} - \displaystyle\int_{a}^{b} (-1)f^2(t)\,dt$
$I = (0 - (b - a)f^2(a)) + \displaystyle\int_{a}^{b} f^2(t)\,dt$
$I = \displaystyle\int_{a}^{b} f^2(t)\,dt - \displaystyle\int_{a}^{b} f^2(a)\,dt$
$I = \displaystyle\int_{a}^{b} (f^2(x) - f^2(a))\,dx$
Answer: $\displaystyle\int_{a}^{b}(f^2(x) - f^2(a))\,dx$
Hence, the correct answer is option: 'A'.