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Chapter 7: Problem 72

Use integration by parts to show that if \(f^{\prime}\) is continuous on \([a, b],\) then $$\int_{a}^{b} f(x) f^{\prime}(x) d x=\frac{1}{2}\left(f(b)^{2}-f(a)^{2}\right).$$

Short Answer

Expert verified
Question: Prove that for any differentiable function f(x), it holds that \(\int_{a}^{b} f(x) f^{\prime}(x) dx = \frac{1}{2}\left(f(b)^{2} - f(a)^{2}\right)\). Solution: Apply integration by parts technique and follow the 5 steps: 1. Choose u = f(x) and dv = f'(x)dx. 2. Compute du = f'(x)dx and v = f(x). 3. Use the formula \(\int u \, dv = uv - \int v \, du\). 4. Plug in u, v, du, and dv, then simplify. 5. Divide by 2 and conclude the proof: \(\int_{a}^{b} f(x) f^{\prime}(x) dx = \frac{1}{2}\left(f(b)^{2} - f(a)^{2}\right)\).

Step by step solution

01

Identify u and dv

To use integration by parts, we need to choose u and dv. In this case, we will choose: $$u = f(x)$$ $$dv = f'(x) dx$$ Using these choices, we can find du and v by differentiating u and integrating dv, respectively.
02

Compute du and v

Differentiate u to find du: $$du = f'(x) dx$$ Integrate dv to find v: $$v = \int f'(x) dx = f(x)$$
03

Apply the integration by parts formula

Now that we have all the necessary components, we can apply the integration by parts formula: $$\int_{a}^{b} u \, dv = \int_{a}^{b} f(x)f'(x)dx = uv\Big|_{a}^{b} - \int_{a}^{b} v \, du$$
04

Plug in u, v, du, and dv and simplify

substituting our chosen u, v, du, and dv into the equation from Step 3: $$\int_{a}^{b} f(x) f^{\prime}(x) d x = f(x)^2 \Big|_{a}^{b} - \int_{a}^{b}f(x)f'(x) dx$$ Simplify the left-hand side: $$\int_{a}^{b} f(x) f^{\prime}(x) d x = f(b)^2 - f(a)^2 - \int_{a}^{b} f(x) f^{\prime}(x) d x$$ Add \(\int_{a}^{b} f(x) f^{\prime}(x) d x\) to both sides: $$2\int_{a}^{b} f(x) f^{\prime}(x) d x = f(b)^2 - f(a)^2$$
05

Divide by 2 and conclude

Divide both sides of the equation by 2: $$\int_{a}^{b} f(x) f^{\prime}(x) d x=\frac{1}{2}\left(f(b)^{2}-f(a)^{2}\right).$$ Hence, we have proven the given statement: $$\int_{a}^{b} f(x) f^{\prime}(x) d x=\frac{1}{2}\left(f(b)^{2}-f(a)^{2}\right).$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a fundamental concept in calculus that involves finding the rate at which a function is changing at any given point. The derivative of a function, denoted as \( f'(x) \), gives us this rate of change and is essentially the slope of the tangent line at any point \( x \) on the curve of the function \( f(x) \).

Understanding differentiation is crucial as it lies at the heart of many advanced calculus topics, such as integration by parts. In the exercise you're tackling, differentiation helps us in setting up the integration by parts formula.
  • We differentiate \( u \), which is equal to \( f(x) \), to find \( du = f'(x) dx \).
  • This derivative represents small changes in \( f \) with respect to \( x \), a pivotal step when applying the integration by parts technique.
Differentiation connects to the bigger picture of rate of change, motion, and optimization in calculus; allowing us to navigate through function behaviors and their tendencies at specific points.
Continuity
Continuity describes how functions behave and provide an unbroken, smooth progression of values. A function \( f(x) \) is continuous on an interval \([a, b]\) if there are no breaks, jumps, or holes within that stretch. Specifically, for the derivative \( f'(x) \) to be continuous over \([a, b]\), it follows that \( f(x) \) must not suddenly jump or deviate sharply, making it predictable and consistent.
  • One important implication of continuity for derivatives like \( f'(x) \) is that it assures the limits and integrals are well-defined over the interval \([a, b]\).

  • For integration by parts, we rely on the smooth behavior ensured by continuity, as breaking continuity could lead to undefined or infinite results.
Contemporary calculus heavily leans on concepts of continuity to ensure functions can be properly analyzed and processed further, like in establishing the integral relationships necessary for solving by parts.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links the concept of differentiation and integration, serving as a cornerstone of calculus. It states that if a function is continuous over an interval \([a, b]\), and \( F \) is an antiderivative of \( f \) over \([a, b]\), then

\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]

This theorem underpins the process of evaluating definite integrals, using antiderivatives evaluated at the upper and lower bounds of the interval.
In the context of our problem, the theorem justifies our computation where we integrate \( f'(x) \) to retrieve \( f(x) \), intertwining it with the integration by parts formula.
  • This blend of derivative and integral highlights how calculus elegantly combines these concepts to solve broader problems.

  • The continuity of \( f'(x) \) plays a role here, ensuring that the theorem is applicable without exceptions or discontinuities over \([a, b]\).
Thus, the Fundamental Theorem of Calculus seamlessly combines differentiation and integration, proving essential in the working out of calculus problems like the one with integration by parts.

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Most popular questions from this chapter

Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}[\cos ((m-n) x)-\cos ((m+n) x)]\\\&\sin m x \cos n x=\frac{1}{2}[\sin ((m-n) x)+\sin ((m+n) x)]\\\&\cos m x \cos n x=\frac{1}{2}[\cos ((m-n) x)+\cos ((m+n) x)]\end{aligned}$$ $$\int \cos x \cos 2 x d x$$

The nucleus of an atom is positively charged because it consists of positively charged protons and uncharged neutrons. To bring a free proton toward a nucleus, a repulsive force \(F(r)=k q Q / r^{2}\) must be overcome, where \(q=1.6 \times 10^{-19} \mathrm{C}\) is the charge on the proton, \(k=9 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}, Q\) is the charge on the nucleus, and \(r\) is the distance between the center of the nucleus and the proton. Find the work required to bring a free proton (assumed to be a point mass) from a large distance \((r \rightarrow \infty)\) to the edge of a nucleus that has a charge \(Q=50 q\) and a radius of \(6 \times 10^{-11} \mathrm{m}.\)

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The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{x^{4}+1}{x^{3}+9 x} d x$$

Let \(I_{n}=\int x^{n} e^{-x^{2}} d x,\) where \(n\) is a nonnegative integer. a. \(I_{0}=\int e^{-x^{2}} d x\) cannot be expressed in terms of elementary functions. Evaluate \(I_{1}\). b. Use integration by parts to evaluate \(I_{3}\). c. Use integration by parts and the result of part (b) to evaluate \(I_{5}\). d. Show that, in general, if \(n\) is odd, then \(I_{n}=-\frac{1}{2} e^{-x^{2}} p_{n-1}(x)\) where \(p_{n-1}\) is a polynomial of degree \(n-1\). e. Argue that if \(n\) is even, then \(I_{n}\) cannot be expressed in terms of elementary functions.
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