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Chapter 5: Problem 104

Substitution: shift Perhaps the simplest change of variables is the shift or translation given by \(u=x+c,\) where \(c\) is a real number. a. Prove that shifting a function does not change the net area under the curve, in the sense that $$\int_{a}^{b} f(x+c) d x=\int_{a+c}^{b+c} f(u) d u$$ b. Draw a picture to illustrate this change of variables in the case that \(f(x)=\sin x, a=0, b=\pi,\) and \(c=\pi / 2\)

Short Answer

Expert verified
Question: Prove that shifting a function does not change the net area under the curve, and provide an example to illustrate the change of variables. Answer: By substituting u = x + c in the given integral equation, we showed that: $$\int_a^b f(x + c) dx = \int_{a+c}^{b+c} f(u) du$$ This means that the net area under the curve remains the same when a function is shifted. We illustrated this by using the example of f(x) = sin x with a = 0, b = 蟺, and c = 蟺 / 2: $$\int_0^{\pi} sin(x + \frac{\pi}{2}) dx = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} sin(u) du$$ Both integrals have the same net area under the curve, proving that shifting a function does not change the net area under the curve.

Step by step solution

01

Substitute u in the given integral equation

First, we start by substituting u = x + c in the given integral equation: $$\int_a^b f(x + c) dx = \int_{a+c}^{b+c} f(u) du$$
02

Perform the substitution

Now, we perform the substitution: \(x = u - c\), and \(dx = du\) $$\int_a^b f(u - c) (d(u - c)) = \int_{a+c}^{b+c} f(u) du$$
03

Adjust the integral limits

Now, adjust the integral limits: When \(x=a\), then \(u = a+c\), and when \(x=b\), then \(u = b+c\). So, our substituted integral becomes: $$\int_{a+c}^{b+c} f(u - c) du = \int_{a+c}^{b+c} f(u) du$$
04

Comparing both integrals

Comparing both the integrals, we can see that they are equal. $$\int_a^b f(x + c) dx = \int_{a+c}^{b+c} f(u) du$$
05

Illustrating with an example

Now, we will illustrate this change of variables with f(x) = sin x, a = 0, b = 蟺, and c = 蟺 / 2. Let \(u = x + \frac{\pi}{2}\), so \(f(u) = sin(u - \frac{\pi}{2})\). The integral equation becomes: $$\int_0^{\pi} sin(x + \frac{\pi}{2}) dx = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} sin(u) du$$ The left side of the equation represents the integral of sin(x + 蟺/2) from 0 to 蟺, whereas the right side of the equation represents the integral of sin(u) from 蟺/2 to 3蟺/2. You can draw the graphs of both functions side by side, and visually verify that the net area under the curve is the same for both integrals.

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