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Chapter 2: Problem 14

Given the function \(f(x)=\left\\{\begin{array}{l}x^{2} \text { for } 0 \leq x<1 \\ \sqrt{x} \text { for } 1 \leq x \leq 2\end{array}\right.\) Compute \(\int_{0}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx}\).

Short Answer

Expert verified
Answer: The value of the integral $\int_{0}^{2} f(x) dx$ for the given piecewise-defined function is $\frac{4\sqrt{2} - 1}{3}$.

Step by step solution

01

Split the integral into two parts

Since the function is defined differently on the intervals \([0,1)\) and \([1,2],\) we can split the integral as follows: \(\int_{0}^{2} f(x) dx = \int_{0}^{1} x^2 dx + \int_{1}^{2} \sqrt{x} dx\)
02

Evaluate the first integral

To compute the first integral, we evaluate the antiderivative of the function \(x^2\): \(\int x^2 dx =\frac{1}{3}x^3 + C\) Now find the definite integral on the interval \([0, 1]\): \(\int_{0}^{1} x^2 dx = \left[\frac{1}{3}(1)^3 - \frac{1}{3}(0)^3\right]\) \(= \left[\frac{1}{3}(1) - \frac{1}{3}(0)\right] = \frac{1}{3}\)
03

Evaluate the second integral

For the second integral, we'll first rewrite the function as a power of x, which then we can evaluate the antiderivative: \(\int \sqrt{x} dx = \int x^{\frac{1}{2}} dx = \frac{2}{3} x^{\frac{3}{2}} + C\) Now find the definite integral on the interval \([1, 2]\): \(\int_{1}^{2} \sqrt{x} dx = \left[\frac{2}{3}(2)^{\frac{3}{2}} - \frac{2}{3}(1)^{\frac{3}{2}}\right]\) \(= \left[\frac{2}{3}(2\sqrt{2}) - \frac{2}{3}(1)\right] = \frac{2}{3}(2\sqrt{2} - 1)\)
04

Add the results of the two integrals

Now we'll combine the results from Steps 2 and 3 to get the final value of the integral: \(\int_{0}^{2} f(x) dx = \int_{0}^{1} x^2 dx + \int_{1}^{2} \sqrt{x} dx\) \(= \frac{1}{3} + \frac{2}{3}(2\sqrt{2} - 1)\) \(= \frac{1}{3} + \frac{4\sqrt{2}}{3} - \frac{2}{3}\) \(= \frac{4\sqrt{2} - 1}{3}\) Therefore, the integral \(\int_{0}^{2} f(x) dx = \frac{4\sqrt{2} - 1}{3}.\)

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Most popular questions from this chapter

(a) Show that \(\int_{-1}^{1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\mathrm{f}\left(\frac{-1}{\sqrt{3}}\right)+\mathrm{f}\left(\frac{1}{\sqrt{3}}\right)\) for \(f(x)=1, x, x^{2}\) and \(x^{3}\) (b) Let a and b be two numbers, \(-1 \leq \mathrm{a}<\mathrm{b} \leq 1\) such that \(\int_{-1}^{1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\mathrm{f}(\mathrm{a})+\mathrm{f}(\mathrm{b})\) for \(\mathrm{f}(\mathrm{x})=1\), \(x, x^{2}\), and \(x^{3} .\) Show that \(a=-1 / \sqrt{3}\) and \(b=1 / \sqrt{3}\). (c) Show that the approximation \(\int_{-1}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx} \approx \mathrm{f}(-1 / \sqrt{3})+\mathrm{f}(1 / \sqrt{3})\) has no error when \(\mathrm{f}\) is a polynomial of degree atmost 3 .

(a) Make a conjecture about the value of the limit \(\lim _{k \rightarrow 0} \int_{1}^{b} t^{k-1} d t(b>0)\) (b) Check your conjecture by evaluating the integral and finding the limit. [Hint: Interpret the limit as the definition of the derivative of an exponential function]

Show that \(\int_{0}^{\infty} \sin \theta \mathrm{d} \theta\) and \(\int_{0}^{\infty} \cos \theta \mathrm{d} \theta\) are indeterminate.

Showthat \(\int_{0}^{\pi} \frac{\ell \mathrm{n}(1+\mathrm{a} \cos \mathrm{x})}{\cos \mathrm{x}} \mathrm{dx}=\pi \sin ^{-1} \mathrm{a},(|\mathrm{a}|<1)\)

Find \(\int_{0}^{2} f(x) d x\), where \(f(x)=\left\\{\begin{array}{c}\frac{1}{\sqrt[4]{x^{3}}} \quad \text { for } 0 \leq x \leq 1 \\ \frac{1}{\sqrt[4]{(x-1)^{3}}}\end{array}\right.\) for \(1
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