Problem 5
Question
Prove integration by parts. That is, suppose \(F\) and \(G\) are continuously differentiable functions on \([a, b] .\) Then prove $$\int_{a}^{b} F(x) G^{\prime}(x) d x=F(b) G(b)-F(a) G(a)-\int_{a}^{b} F^{\prime}(x) G(x) d x.$$
Step-by-Step Solution
Verified Answer
Integration by parts is derived from the product rule, rearranging gives the formula.
1Step 1: State the Integration by Parts Formula
Integration by parts is a technique derived from the product rule for differentiation. It is expressed as:\[ \int F(x) G'(x) \, dx = F(x)G(x) - \int G(x)F'(x) \, dx \]}
2Step 2: Recall the Product Rule for Differentiation
The product rule for differentiation states:\[ \frac{d}{dx} [F(x)G(x)] = F(x)G'(x) + G(x)F'(x) \].This will be the basis for deriving the integration by parts formula.
3Step 3: Integrate the Product Rule over [a, b]
Integrate both sides of the product rule equation from \(a\) to \(b\):\[ \int_{a}^{b} \frac{d}{dx}[F(x)G(x)] \, dx = \int_a^b [F(x)G'(x) + G(x)F'(x)] \, dx \].
4Step 4: Apply the Fundamental Theorem of Calculus
By the Fundamental Theorem of Calculus, the left side simplifies to the evaluation of \([F(x)G(x)]\) at the boundaries \(a\) and \(b\).\[ F(b)G(b) - F(a)G(a) = \int_a^b F(x)G'(x) \, dx + \int_a^b G(x)F'(x) \, dx \].
5Step 5: Rearrange the Equation
Rearrange the equation to isolate the integral with \(F(x)G'(x)\):\[ \int_a^b F(x)G'(x) \, dx = F(b)G(b) - F(a)G(a) - \int_a^b G(x)F'(x) \, dx \].This is the integration by parts formula.
Key Concepts
Product Rule for DifferentiationFundamental Theorem of CalculusContinuously Differentiable Functions
Product Rule for Differentiation
When we think about the derivative of a product of two functions, we use the product rule. This is a fundamental concept in calculus. The product rule is formulated as follows: \[ \frac{d}{dx} [F(x)G(x)] = F(x)G'(x) + G(x)F'(x) \] Simply put, it tells us how to differentiate two functions that are multiplied together. Here's how it works:
- Consider two functions, \(F(x)\) and \(G(x)\).
- The derivative of their product is not simply the product of their derivatives.
- Instead, it involves each function and the derivative of the other function.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects differentiation and integration, serving as a bridge between the two central operations in calculus. The theorem comes in two parts:
- The first part states that if \(F\) is an antiderivative of \(f\), then the integral of \(f(x)\) from \(a\) to \(b\) is \(F(b) - F(a)\).
- The second part asserts that if \(f\) is continuous on \([a, b]\), and \(F\) is defined by the integral of \(f\) from \(a\) to \(x\), then \(F'(x) = f(x)\).
Continuously Differentiable Functions
Continuously differentiable functions are those that not only have derivatives, but do so smoothly and without interruption across their domain. In calculus, a function is said to be continuously differentiable over an interval if both the function and its derivative are continuous within that interval. Here are some key points:
- A function \(F(x)\) is continuously differentiable on \([a, b]\) if \(F'(x)\) exists and is continuous on \([a, b]\).
- It implies that there are no sudden jumps or breaks in the slope of the function.
- This smoothness is crucial for many calculus techniques, including integration by parts.
Other exercises in this chapter
Problem 5
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