/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 (a) Let \(f(x):=2\) if \(0 \leq ... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 6

(a) Let \(f(x):=2\) if \(0 \leq x<1\) and \(f(x):=1\) if \(1 \leq x \leq 2\). Show that \(f \in \mathcal{R}[0,2]\) and evaluate its integral. (b) Let \(h(x):=2\) if \(0 \leq x<1, h(1):=3\) and \(h(x):=1\) if \(1

Short Answer

Expert verified
Both f(x) and h(x) are Riemann-integrable over [0,2] and their integrals are equal to 3.

Step by step solution

01

Verify Riemann Integrability for f(x)

As per the definition, f(x) is well-defined and bounded over the closed interval [0,2]. Hence, f(x) will be Riemann-integrable.
02

Calculate Integral of f(x)

The definite integral of f(x) from 0 to 2 is equal to the sum of the areas of the rectangles under the graph. This can be calculated as:\[\int_0^1 f(x) dx + \int_1^2 f(x) dx = \int_0^1 2 dx + \int_1^2 1 dx = [2x]_0^1 + [x]_1^2 = 2*(1-0) + (2-1) = 2+1 = 3\]
03

Verify Riemann Integrability for h(x)

Similar to f(x), h(x) is also well-defined and bounded over the closed interval [0,2]. Hence, h(x) is also Riemann-integrable.
04

Calculate Integral of h(x)

The definite integral of h(x) from 0 to 2 is similar to part a:\[\int_0^1 h(x) dx + \int_1^2 h(x) dx = \int_0^1 2 dx + \int_1^2 1 dx = [2x]_0^1 + [x]_1^2 = 2*(1-0) + (2-1) = 2+1 = 3\]The only difference is that, at x=1, h(x) is equal to 3, but this doesn't affect the integral since the set of discontinuities has a zero measure in [0,2].

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Most popular questions from this chapter

If \(g(x):=x\) for \(|x| \geq 1\) and \(g(x):=-x\) for \(|x|<1\) and if \(G(x):=\frac{1}{2}\left|x^{2}-1\right|\), show that \(\int_{-2}^{3} g(x) d x=G(3)-G(-2)=5 / 2\)

If \(g \in \mathcal{R}[a, b]\) and if \(f(x)=g(x)\) except for a finite number of points in \([a, b]\), prove that \(f \in \mathcal{R}[a, b]\) and that \(\int_{a}^{b} f=\int_{a}^{b} g\)

If \(f\) is bounded by \(M\) on \([a, b]\) and if the restriction of \(f\) to every interval \([c, b]\) where \(c \in(a, b)\) is Riemann integrable, show that \(f \in \mathcal{R}[a, b]\) and that \(\int_{c}^{b} f \rightarrow \int_{a}^{b} f\) as \(c \rightarrow a+\). [Hint: Let \(\alpha_{c}(x):=-M\) and \(\omega_{c}(x):=M\) for \(x \in[a, c)\) and \(\alpha_{c}(x):=\omega_{c}(x):=f(x)\) for \(x \in[c, b]\). Apply the Squeeze Theorem \(7.2 .3\) for \(c\) sufficiently near \(a\).]

Let \(f, g \in \mathcal{R}[a, b\\}\) (a) If \(t \in \mathbb{R}\), show that \(\int_{a}^{b}(t f \pm g)^{2} \geq 0\). (b) Use (a) to show that \(2\left|\int_{a}^{b} f g\right| \leq t \int_{a}^{b} f^{2}+(1 / t) \int_{a}^{b} g^{2}\) for \(t>0\). (c) If \(\int_{a}^{b} f^{2}=0\), show that \(\int_{a}^{b} f g=0\). (d) Now prove that \(\left|\int_{a}^{b} f g\right|^{2} \leq\left(\int_{a}^{b}|f g|\right)^{2} \leq\left(\int_{a}^{b} f^{2}\right) \cdot\left(\int_{a}^{b} g^{2}\right) .\) This inequality is called the Cauchy-Bunyakovsky-Schwarz Inequality (or simply the Schwarz Inequality).

Approximate the indicated integrals, giving estimates for the error. Use a calculator to obtain a high degree of precision.$$ \int_{0}^{2}\left(1+x^{4}\right)^{1 / 2} d x $$
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