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Chapter 18: Problem 24

Prove that $$\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x.$$ (A geometric interpretation makes this clear, but it is also a good exercise in the handling of limits of integration during a substitution.)

Short Answer

Expert verified
\( \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx \) by substitution \( u = a + b - x \).

Step by step solution

01

Define the substitution

Let鈥檚 perform the substitution. Define a new variable of integration by setting \[ u = a + b - x. \]
02

Determine the differential

Find the differential of the new variable of integration:\[ du = -dx \]
03

Adjust integration limits

Change the limits of integration according to the substitution. When \( x = a \), \[ u = a + b - a = b, \] and when \( x = b \), \[ u = a + b - b = a. \] So, the new integration limits are from \( u = b \) to \( u = a \).
04

Rewrite the integral

Substitute everything into the integral. The integral becomes: \[ \int_{a}^{b} f(x) \, dx = \int_{b}^{a} f(u) \, (-du) \]
05

Adjust limits and sign

Changing the order of integration limits swaps their signs. Therefore, we have\[ \int_{b}^{a} f(u) \, (-du) = \int_{a}^{b} f(u) \, du \]
06

Substitute back to original variable

Recall that \( u = a + b - x \), so replacing \( u \) with \( a + b - x \) gives: \[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \]

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Key Concepts

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

Integration Limits
When evaluating integrals, the limits of integration indicate the interval over which we are integrating. For a definite integral \(\text {\int_{a}^{b} f(x)dx}\), these limits are particularly crucial.
They show us the starting point \(\text {a}\) and the endpoint \(\text {b}\) of the integral on the x-axis.
Changing the variable of integration, as seen in substituting \(\text {u = a + b - x}\), affects these limits directly. Integrating from \(\text {a}\) to \(\text {b}\) with respect to \(\text {x}\) becomes integrating from \(\text {b}\) to \(\text {a}\) with respect to \(\text {u}\).
Don't worry if this seems confusing. Remember, the limits switch places according to the new variable of integration.
Variable Substitution
Variable substitution, also known as u-substitution, is a powerful technique in calculus for simplifying integrals.
By substituting a new variable, we can transform a complex integral into a simpler one. For example, in the given problem, we set \(\text {u = a + b - x}\) to make the integral more manageable.
Always remember to change the differential accordingly; here, \(\text {du = -dx}\). Substitution often requires adjusting the integration limits to correspond to the new variable.
  • Helps in simplifying complex integrals
  • Requires appropriate transformation of limits and differentials
Differential Calculus
Differential calculus focuses on the concept of a derivative, which represents the rate of change. In the context of this integral problem, finding the differential, as in \(\text {du = -dx}\), is critical.
This change in differential is essential for properly rewriting the integral in terms of the new variable.
Key Points:
  • Differentiation gives the rate of change
  • Finding differentials helps in substitution processes
The substitution \(\text {u = a + b - x}\) results directly in \(\text {-dx}\), helping us adjust the integral correctly during transformation.
Definite Integrals
A definite integral computes the accumulated value over an interval [a, b]. For the expression \(\text {\int_{a}^{b} f(x)dx}\), it evaluates the net area under the curve \(\text {f(x)}\) from \(\text {x = a}\) to \(\text {x = b}\).
In the given problem, after substitution, the integral transforms but still represents the same area, thus proving the equality.
How It Works:
  • Evaluates the accumulated value over an interval
  • Changing variables does not change the integral's value but simplifies the problem
By reworking the integral's limits and variables, we retain the meaning and value of the integral, showing the equivalence between \(\text { \int_{a}^{b} f(x)dx }\) and \(\text {\int_{a}^{b} f(a+b-x)dx }\).

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Most popular questions from this chapter

(a) Show that the following improper integrals both converge. (i) \(\int_{0}^{1} \sin \left(x+\frac{1}{x}\right) d x\). (ii) \(\int_{0}^{1} \sin ^{2}\left(x+\frac{1}{x}\right) d x\). (b) Decide which of the following improper integrals converge. (i) \(\int_{1}^{\infty} \sin \left(\frac{1}{x}\right) d x\). (ii) \(\int_{1}^{\infty} \sin ^{2}\left(\frac{1}{x}\right) d x\).

Derive the formula for \(\int \sec x \, d x\) in the following two ways: (a) By writing $$\begin{aligned} \frac{1}{\cos x} &=\frac{\cos x}{\cos ^{2} x} \\ &=\frac{\cos x}{1-\sin ^{2} x} \\ &=\frac{1}{2}\left[\frac{\cos x}{1+\sin x}+\frac{\cos x}{1-\sin x}\right], \end{aligned}$$ an expression obviously inspired by partial fraction decompositions. Be sure to note that \(\int \cos x /(1-\sin x) d x=-\log (1-\sin x) ;\) the minus sign is very important. And remember that \(\frac{1}{2} \log \alpha=\log \sqrt{\alpha} .\) From there on, keep doing algebra, and trust to luck. (b) By using the substitution \(t=\tan x / 2 .\) One again, quite a bit of manipulation is required to put the answer in the desired form; the expression \(\tan x / 2\) can be attacked by using Problem \(15-9,\) or both answers can be expressed in terms of \(t .\) There is another expression for \(\int \sec x \, d x\) which is less cumbersome than \(\log (\sec x+\tan x) ;\) using Problem 15-9 we obtain $$\int \sec x \, d x=\log \left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)=\log \left(\tan \left(\frac{x}{2}+\frac{\pi}{4}\right)\right).$$ This last expression was actually the one first discovered. and was due, not to any mathematician's cleverness, but to a curious historical accident: In 1599 Wright computed nautical tables that amounted to definite integrals of sec. When the first tables for the logarithms of tangents were produced, the correspondence between the two tables was immediately noticed (but remained unexplained until the invention of calculus).

The derivation of \(\int e^{x} \sin x d x\) given in the text seems to prove that the only primitive of \(f(x)=e^{x} \sin x\) is \(F(x)=e^{x}(\sin x-\cos x) / 2,\) whereas \(F(x)=\) \(e^{x}(\sin x-\cos x) / 2+C\) is also a primitive for any number \(C .\) Where does \(C\) come from? (What is the meaning of the equation $$\int e^{x} \sin x d x=e^{x} \sin x-e^{x} \cos x-\int e^{x} \sin x d x ?$$

The following integrations can all be done with substitutions of the form \(x=\sin u, x=\cos u,\) etc. To do some of these you will need to remember that $$\int \sec x \, d x=\log (\sec x+\tan x)$$ as well as the following formula, which can also be checked by differentiation: $$\int \csc x \, d x=-\log (\csc x+\cot x).$$ In addition, at this point the derivatives of all the trigonometric functions should be kept handy. (i) \(\int \frac{d x}{\sqrt{1-x^{2}}} \cdot\) (You already know this integral, but use the substitution \(x=\sin u\) anyway, just to see how it works out.) (ii) \(\int \frac{d x}{\sqrt{1+x^{2}}}\) (Since \(\tan ^{2} u+1=\sec ^{2} u,\) you want to use the substitution \(x=\tan u .)\) (iii) \(\int \frac{d x}{\sqrt{x^{2}-1}}\). (iv) \(\int \frac{d x}{x \sqrt{x^{2}-1}}\) (The answer will be a certain inverse function that was given short shrift in the text.) (v) \(\int \frac{d x}{x \sqrt{1-x^{2}}}\). (vi) \(\int \frac{d x}{x \sqrt{1+x^{2}}}\). \(\left.\begin{array}{l}\text { (vii) } \int x^{3} \sqrt{1-x^{2}} d x \\ \text { (viii) } \int \sqrt{1-x^{2}} d x\end{array}\right\\} \begin{array}{l}\text { You will need to remember the methods for } \\ \text { integrating powers of sin and cos. }\end{array}.\) (ix) \(\int \sqrt{1+x^{2}} d x\). \((x) \quad \int \sqrt{x^{2}-1} d x\).

Prove the following version of integration by parts for improper integrals: $$\int_{a}^{\infty} u^{\prime}(x) v(x) d x=\left.u(x) v(x)\right|_{a} ^{\infty}-\int_{a}^{\infty} u(x) v^{\prime}(x) d x.$$ The first symbol on the right side means, of course, $$\lim _{x \rightarrow \infty} u(x) v(x)-u(a) v(a).$$
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