We will use the Fundamental Theorem of Calculus (FTC) to prove the given equation:

$\int_a^b f(x) dx + \int_c^d g(y) dy = bd - ac$

for a continuous, increasing bijection $f: [a,b] \to [c,d]$ with inverse $g = f^{-1}$.

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Step 1: Define the Functions

• Since $f$ is a continuous, increasing bijection, it has a well-defined inverse $g: [c,d] \to [a,b]$, meaning $g(y) = f^{-1}(y)$.
• We treat $b$ as a variable and recognize that $d = f(b)$.

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Step 2: Define a Function $S(b)$

Define the function:

$S(b) = \int_a^b f(x)dx + \int_c^{f(b)} g(y)dy$

We aim to show that $S(b) = bd - ac$, so we differentiate both sides with respect to $b$.

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Step 3: Differentiate Both Sides

Using the Leibniz rule for differentiation under the integral sign:

$\frac{d}{db} \int_a^b f(x) dx = f(b)$

and using the FTC and chain rule for the second integral:

$\frac{d}{db} \int_c^{f(b)} g(y) dy = g(f(b)) \cdot f'(b)$

Since $g = f^{-1}$, we know that $g(f(b)) = b$, so:

$\frac{d}{db} \int_c^{f(b)} g(y) dy = b f'(b)$

Thus, differentiating $S(b)$:

$S'(b) = f(b) + b f'(b)$

Since we want to show $S(b) = bd - ac$, we differentiate the right-hand side:

$\frac{d}{db} (bd - ac) = d + b \cdot f'(b)$

Since $d = f(b)$, we get:

$S'(b) = f(b) + b f'(b) = d + b f'(b)$

Since both derivatives are equal and $S(a) = ac - ac = 0$, we conclude:

$S(b) = bd - ac.$

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Step 4: What If $f$ Is Decreasing?

If $f$ is decreasing, the same proof holds, but we get negative values for the integrals, which still sum to $bd - ac$ as expected.

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Conclusion

Using the Fundamental Theorem of Calculus and differentiating both sides, we have proven:

$\int_a^b f(x) dx + \int_c^d g(y) dy = bd - ac.$

Would you like a more detailed explanation of any step?

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Related Questions

1. How does the Fundamental Theorem of Calculus apply to inverse functions in general?
2. Can this proof be extended to functions that are not bijections but still have an inverse on a subinterval?
3. What happens if $f$ is not continuous? Does the result still hold?
4. How would this proof change if $f$ were a piecewise function with discontinuities?
5. Can we generalize this proof for functions defined on unbounded intervals?

Tip

When working with inverse functions in integrals, always consider changing the variable of integration to simplify the problem.