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Chapter 1: Problem 20

Show that \(\int|x| d x=\frac{x|x|}{2}+C\)

Short Answer

Expert verified
Question: Prove that \(\int|x| dx=\frac{x|x|}{2}+C\). Answer: We have shown that for positive \(x\), the antiderivative is \(\frac{x^2}{2}+C_1\), while for negative \(x\), it is \(-\frac{x^2}{2}+C_2\). To write this in a single line with the absolute value of x, we consider the function \(\frac{x|x|}{2}\). Adding a general constant \(C\) to account for both \(C_1\) and \(C_2\), we obtain the combined antiderivative as \(\int|x| dx=\frac{x|x|}{2}+C\).

Step by step solution

01

Case 1: x > 0

For \(x>0\), the function \(|x|\) is equal to \(x\), so our integral becomes \(\int x dx\). We evaluate this integral to obtain the antiderivative: \(\frac{x^2}{2}+C_1\), where \(C_1\) is a constant.
02

Case 2: x < 0

For \(x<0\), the function \(|x|\) is equal to \(-x\), so our integral becomes \(\int -x dx\). We evaluate this integral to obtain the antiderivative: \(-\frac{x^2}{2}+C_2\), where \(C_2\) is a constant.
03

Combining both cases

We have to combine our results for both cases, and choose a suitable general constant \(C\). For positive \(x\), the antiderivative contains the term \(\frac{x^2}{2}\), while for negative \(x\) it is \(-\frac{x^2}{2}\); we can write this in one line with the absolute value of x in the denominator: \(\frac{x|x|}{2}\). Since \(C_1\) and \(C_2\) are constants, we can represent them with a single constant \(C\). Therefore, the antiderivative for the given exercise is: \(\int|x| dx=\frac{x|x|}{2}+C\)

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Most popular questions from this chapter

Evaluate the following integrals: (i) \(\int \frac{\mathrm{dx}}{\mathrm{x} \sqrt{\left(9 \mathrm{x}^{2}+4 \mathrm{x}+1\right)}}\) (ii) \(\int \frac{d x}{(1+x) \sqrt{\left(1+x-x^{2}\right)}}\) (iii) \(\int \frac{\mathrm{dx}}{(1+\mathrm{x}) \sqrt{1+2 \mathrm{x}-\mathrm{x}^{2}}}\) (iv) \(\int \frac{2 x d x}{\left(1-x^{2}\right) \sqrt{\left(x^{4}-1\right)}}\)

Evaluate \(\int \frac{9 x^{3}-3 x^{2}+2}{\sqrt{3 x^{2}-2 x+1}} d x\)

Only one of the functions \(\sqrt{x} \sqrt[3]{1-x}\) and \(\sqrt{1-x} \sqrt[3]{1-x}\) has an elementary antiderivative. Find the function.

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