Integrals Involving Quadratics — Question 4

Question 4 – Solution

Step 1: Complete the square. $$x^2-4x+8=(x-2)^2+4.$$ Step 2: Find the maximum. The denominator is smallest when $(x-2)^2=0$, or $x=2$. Its minimum is $4$, so $$f(2)=\frac14.$$ The antiderivative below has finite limits at both $-\infty$ and $\infty$, so both one-sided improper integrals converge separately. Their sum may therefore be computed using symmetric limits. Step 3: Write the improper integral as a limit. $$\begin{align*} I&=\lim_{R\to\infty}\int_{-R}^{R}\frac{dx}{(x-2)^2+2^2}\\ &=\lim_{R\to\infty}\frac12 \left[\arctan\left(\frac{x-2}{2}\right)\right]_{-R}^{R}. \end{align*}$$ Step 4: Evaluate the limits. $$\begin{align*} I&=\frac12\left(\frac\pi2-\left(-\frac\pi2\right)\right) =\frac\pi2. \end{align*}$$ $$\boxed{\max=1/4,\quad I=\pi/2}$$