Question

Use limits of definite integrals to calculate each of the improper integrals in Exercises.

${\int }_0^4\frac{2x-4}{x^2-4x+3}dx$

Short answer

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

${\int }_0^4\frac{2x-4}{x^2-4x+3}dx=\infty$

Step-by-step solution

Step 1. Given information. 

The given integral is the following.

${\int }_0^4\frac{2x-4}{x^2-4x+3}dx$

Step 2. Value of integral. 

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

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

localid="1648884400399" ${\int }_0^4\frac{2x-4}{x^2-4x+3}dx={\int }_0^1\frac{2x-4}{x^2-4x+3}dx+{\int }_1^3\frac{2x-4}{x^2-4x+3}dx+{\int }_3^4\frac{2x-4}{x^2-4x+3}dx$

As we know thatlocalid="1648884235145" ${\int }_0^1\frac{1}{x^2}dx$is diverse and integral of functions less than $\frac{1}{x^2}$will be diverse.

and $\frac{2x-4}{x^2-4x+3}>\frac{1}{x^2}$so localid="1648891997358" ${\int }_0^1\frac{2x-4}{x^2-4x+3}dx$is diverse.

So${\int }_0^4\frac{2x-4}{x^2-4x+3}dx$is also diverse.

${\int }_0^4\frac{2x-4}{x^2-4x+3}dx=\infty$