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Chapter 5: Q. 33 (page 478)

Use limits of definite integrals to calculate each of the improper integrals in Exercises.
${鈭�}_{0}^{4} \frac{2 x - 4}{x^{2} - 4 x + 3} d x$

Short Answer

Expert verified

The improper integral on $[0, 4]$ diverges.

${鈭�}_{0}^{4} \frac{2 x - 4}{x^{2} - 4 x + 3} d x = 鈭�$

Step by step solution

01

Step 1. Given information.

The given integral is the following.

${鈭�}_{0}^{4} \frac{2 x - 4}{x^{2} - 4 x + 3} d x$

02

Step 2. Value of integral.

The function $f (x) = \frac{2 x - 4}{x^{2} - 4 x + 3}$ has a vertical asymptote at $x = 1, 3$

solve this improper integral by splitting at $x = 1, 3 .$

localid="1648884400399" ${鈭�}_{0}^{4} \frac{2 x - 4}{x^{2} - 4 x + 3} d x = {鈭�}_{0}^{1} \frac{2 x - 4}{x^{2} - 4 x + 3} d x + {鈭�}_{1}^{3} \frac{2 x - 4}{x^{2} - 4 x + 3} d x + {鈭�}_{3}^{4} \frac{2 x - 4}{x^{2} - 4 x + 3} d x$

As we know thatlocalid="1648884235145" ${鈭�}_{0}^{1} \frac{1}{x^{2}} d x$ is diverse and integral of functions less than $\frac{1}{x^{2}}$ will be diverse.

and $\frac{2 x - 4}{x^{2} - 4 x + 3} > \frac{1}{x^{2}}$ so localid="1648891997358" ${鈭�}_{0}^{1} \frac{2 x - 4}{x^{2} - 4 x + 3} d x$ is diverse.

So ${鈭�}_{0}^{4} \frac{2 x - 4}{x^{2} - 4 x + 3} d x$ is also diverse.

${鈭�}_{0}^{4} \frac{2 x - 4}{x^{2} - 4 x + 3} d x = 鈭�$

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