Question

Let \(f(x)\) be a real valued function which is monotonic and differentiable. Then for any reals \(a\) and \(b\), \[ \int_{f(a)}^{f(b)}2x\{b-f^{-1}(x)\}\,dx= \]

• A. $\int_{a}^{b}(f^{2}(x)-f^{2}(a))dx$
• B. $\int_{a}^{b}(f(x)-f(a))^{2}dx$
• C. $\int_{a}^{b}(bf^{2}(x)-af^{2}(a))dx$
• D. $bf^{2}(b)+f^{-1}(a)$

Hint:

An elegant geometric way to view this problem is to visualize it as finding the area under a curve. Drawing out the bounding box regions created by the function transformations shows that the inverse integral blocks shift and balance each other out, collapsing the areas into this clean difference formula.

Answer 1

The Correct Option is A

Concept: Definite integrals that contain an inverse function term $f^{-1}(x)$ can be simplified by applying a substitution step to transform the variable back into the original function's domain space. Setting $x = f(t)$ maps the limits and differentials directly back to the standard variable tracking space. Step 1: Apply variable substitution to shift into the domain space of \(t\).
Let us consider the given definite integral expression: \[ I = \int_{f(a)}^{f(b)} 2x \{b - f^{-1}(x)\}\,dx \] Let us apply the change-of-variable substitution step: \[ x = f(t) \quad \Rightarrow \quad dx = f'(t)\,dt \] By the definition of inverse functions, this mapping simplifies the inverse term to: $f^{-1}(x) = t$.

Step 2: Map the integration limits into the new variable tracking space.
Let us calculate the new integration boundaries using our substitution rules:
• For the lower bound: $x = f(a) \implies f(t) = f(a) \implies t = a$
• For the upper bound: $x = f(b) \implies f(t) = f(b) \implies t = b$ Substitute these new boundaries, terms, and differentials back into our primary integral expression: \[ I = \int_{a}^{b} 2f(t) \{b - t\} \cdot f'(t)\,dt \] Rearrange the terms to group the derivative next to its base function: \[ I = \int_{a}^{b} (b - t) \cdot \Big( 2f(t)f'(t) \Big) \,dt \quad \cdots (1) \]

Step 3: Apply Integration by Parts to solve the expression.
Notice that the second group inside the integrand is the exact derivative of the squared function: $\frac{d}{dt}[f(t)]^2 = 2f(t)f'(t)$. Let us evaluate equation (1) by applying integration by parts ($\int u \cdot v' dt = u \cdot v - \int u' \cdot v dt$):
• Let $u = b - t \implies du = -1 \cdot dt$
• Let $v' = 2f(t)f'(t) \implies v = [f(t)]^2$ Applying the parts formula: \[ I = \Big[ (b - t) \cdot [f(t)]^2 \Big]_{a}^{b} - \int_{a}^{b} (-1) \cdot [f(t)]^2 \,dt \] Substitute the limits into the first bracketed term: at the upper bound $t = b$, the term $(b - b)$ becomes 0, causing that whole piece to drop out: \[ I = \left( 0 - (b - a)[f(a)]^2 \right) + \int_{a}^{b} [f(t)]^2 \,dt \] \[ I = \int_{a}^{b} [f(t)]^2 \,dt - (b - a)[f(a)]^2 \quad \cdots (2) \]

Step 4: Convert the constant boundary subtraction into an integral form.
We can rewrite the constant term $(b-a)[f(a)]^2$ as a simple definite integral of a constant over the same interval boundaries: \[ (b - a)[f(a)]^2 = \int_{a}^{b} [f(a)]^2 \,dt \] Substitute this integral representation back into equation (2) and combine the terms under a single integral sign: \[ I = \int_{a}^{b} [f(t)]^2 \,dt - \int_{a}^{b} [f(a)]^2 \,dt = \int_{a}^{b} \left( [f(t)]^2 - [f(a)]^2 \right) dt \] Swapping the integration variable notation name from $t back to $x gives: \[ I = \int_{a}^{b} \left( f^2(x) - f^2(a) \right) dx \] This matches option (A) perfectly.