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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.3.127

For Exercises 127 and 128 find a function f satisfying each equation.
127. ∫₂ˣ √(f(t)) dt = x ln x

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1
Recognize that the problem gives an integral equation: \(\int_{2}^{x} \sqrt{f(t)} \, dt = x \ln x\). Our goal is to find the function \(f(x)\) that satisfies this equation.
Apply the Fundamental Theorem of Calculus by differentiating both sides of the equation with respect to \(x\). This gives: \(\frac{d}{dx} \int_{2}^{x} \sqrt{f(t)} \, dt = \frac{d}{dx} (x \ln x)\).
By the Fundamental Theorem of Calculus, the derivative of the integral with variable upper limit is the integrand evaluated at \(x\): \(\sqrt{f(x)} = \frac{d}{dx} (x \ln x)\).
Compute the derivative on the right side using the product rule: \(\frac{d}{dx} (x \ln x) = \ln x + 1\).
Set \(\sqrt{f(x)} = \ln x + 1\) and solve for \(f(x)\) by squaring both sides: \(f(x) = (\ln x + 1)^2\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Fundamental Theorem of Calculus

This theorem links differentiation and integration, stating that if F(x) is defined as an integral of f(t) from a constant to x, then F'(x) = f(x). It allows us to find the original function inside an integral by differentiating the integral expression.
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