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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
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6장, 문제 6.1.72

Without evaluating integrals, prove that ∫₀² d/dx(12 sin πx²) dx=∫₀² d/dx (x¹⁰(2−x)³) dx.

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Recognize that both integrals are definite integrals of derivatives over the interval from 0 to 2. According to the Fundamental Theorem of Calculus, for a function \(F(x)\), we have \(\int_a^b F'(x) \, dx = F(b) - F(a)\).
Apply the Fundamental Theorem of Calculus to the first integral: \(\int_0^2 \frac{d}{dx} \left(12 \sin \pi x^2 \right) \, dx = 12 \sin \pi (2)^2 - 12 \sin \pi (0)^2\).
Similarly, apply the Fundamental Theorem of Calculus to the second integral: \(\int_0^2 \frac{d}{dx} \left(x^{10} (2 - x)^3 \right) \, dx = (2)^{10} (2 - 2)^3 - (0)^{10} (2 - 0)^3\).
Evaluate the boundary terms in both expressions without simplifying the trigonometric or polynomial values, just observe their forms: For the first, \(12 \sin (4 \pi)\) and \(12 \sin 0\); for the second, \((2)^{10} \cdot 0^3\) and \(0^{10} \cdot 2^3\).
Note that \(\sin (4 \pi) = 0\) and \(\sin 0 = 0\), and that any term multiplied by zero is zero, so both expressions evaluate to zero, proving the equality of the two integrals without explicitly evaluating the integrals themselves.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Fundamental Theorem of Calculus

This theorem connects differentiation and integration, stating that the integral of a derivative over an interval equals the difference of the original function's values at the interval endpoints. It allows evaluating definite integrals of derivatives without performing integration.
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Properties of Definite Integrals

Definite integrals have linearity and additivity properties, and when integrating a derivative, the integral depends only on the boundary values of the original function. This means integrals of derivatives can be compared by evaluating the functions at the limits.
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Equality of Functions at Interval Endpoints

To prove two integrals of derivatives are equal without integration, it suffices to show the original functions have the same values at the interval's endpoints. If f(a) = g(a) and f(b) = g(b), then ∫ₐᵇ f'(x) dx = ∫ₐᵇ g'(x) dx.
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