smc-7422-60-Flawed Proofs

The following argument attempts to prove that \(0=1\).

We evaluate \(\displaystyle\int \frac{1}{x}\, d x\) using the method of integration by parts.

	\(\displaystyle \int \frac{1}{x}\,d x\)	\(=\displaystyle \int \frac{1}{x} \times 1\, d x\)
		\(=\displaystyle\frac{1}{x} \times x-\int-\frac{1}{x^2} x\, d x\)
		\(=1+\displaystyle\int \frac{1}{x}\, d x\)

We may now subtract \(\displaystyle \int \frac{1}{x}\,d x\) from both sides to show that \(0=1\).

Explain what is wrong with this argument. (2 marks)

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\(\text {Consider the line in proof }\)

\(\displaystyle {\int \frac{1}{x}\, d x=1+\int \frac{1}{x}\, d x}\)

\(\Rightarrow \text{\(\ \)Each integral will have its own constant}\)

\(\Rightarrow \text{ If constants are \(c_1\) and \(c_2, \abs{c_1-c_2}=1\) in this proof.}\)

\(\Rightarrow \text{ Subtracting}\ \displaystyle \int \frac{1}{x}\,d x\ \text{from both sides invalidates the proof.}\)

Show Worked Solution

\(\text {Consider the line in proof }\)

\(\displaystyle {\int \frac{1}{x}\, d x=1+\int \frac{1}{x}\, d x}\)

♦♦ Mean mark 34%.

\(\Rightarrow \text{\(\ \)Each integral will have its own constant}\)

\(\Rightarrow \text{ If constants are \(c_1\) and \(c_2, \abs{c_1-c_2}=1\) in this proof.}\)

\(\Rightarrow \text{ Subtracting}\ \displaystyle \int \frac{1}{x}\,d x\ \text{from both sides invalidates the proof.}\)

Proof, EXT2 P1 2024 HSC 14d