7.4.32 Evaluate the integral by completing the square

In summary, the integral $\displaystyle \int_0^1\dfrac{x}{x^2+4x+13}\, dx$ evaluates to approximately 0.03111 with the completed square method and substitution of $u=x+2$.
  • #1
karush
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7.4.32 Evaluate the integral
$\displaystyle \int_0^1\dfrac{x}{x^2+4x+13}\, dx$
ok side work to complete the square
$x^2+4x=-13$
add 4 to both sides
$x^2+4x+4=-13+4$
simplify
$(x+2)^2+9=0$
ok now whatW|A returned ≈0.03111
 
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  • #2
karush said:
7.4.32 Evaluate the integral
$\displaystyle \int_0^1\dfrac{x}{x^2+4x+13}\, dx$
ok side work to complete the square
$x^2+4x=-13$
add 4 to both sides
$x^2+4x+4=-13+4$
simplify
$(x+2)^2+9=0$
ok now what

$u = x+2 \implies du = dx$

\(\displaystyle \int_0^1 \dfrac{(x+2)-2}{(x+2)^2 + 9} \, dx\)

substitute and reset limits of integration ...

\(\displaystyle \int_2^3 \dfrac{u}{u^2 + 9} - \dfrac{2}{u^2+9} \, du\)

looks like the result will be a log and an inverse tangent ...
 
  • #3
ok little ? about where $u=x+2$ and the limits work...

$$\Biggr|\dfrac{1}{2} \ln(u^2 + 9)-\frac{2}{3}\arctan \left(\frac{u}{3}\right)\Biggr|_2^3$$
 
  • #4
karush said:
ok little ? about where $u=x+2$ and the limits work...

$\Bigg[\dfrac{1}{2} \ln(u^2 + 9)-\frac{2}{3}\arctan \left(\frac{u}{3}\right)\Bigg]_2^3$

$u=x+2$

lower limit of integration is $x=0 \implies x+2 = 0+2 = 2$, the lower limit of integration reset to a $u$ value

do the same for the upper limit of integration

$\Bigg[\dfrac{1}{2} \ln(u^2 + 9)-\dfrac{2}{3}\arctan \left(\dfrac{u}{3}\right)\Bigg]_2^3$

$\left[\dfrac{1}{2}\ln(18)-\dfrac{2}{3}\arctan(1)\right] - \left[\dfrac{1}{2}\ln(13) - \dfrac{2}{3}\arctan\left(\dfrac{2}{3}\right)\right]$

$\ln{\sqrt{\dfrac{18}{13}}} - \dfrac{\pi}{6} + \dfrac{2}{3}\arctan\left(\dfrac{2}{3} \right) \approx 0.03111$
 

Related to 7.4.32 Evaluate the integral by completing the square

1. What is the purpose of completing the square in evaluating an integral?

Completing the square allows us to rewrite the integrand in a form that is easier to integrate, making the evaluation process more manageable.

2. How do you complete the square in an integral?

To complete the square, we take the coefficient of the squared term, divide it by 2, and then square that result. We then add and subtract this value inside the parentheses of the integrand.

3. Why is it important to evaluate integrals using different methods?

Evaluating integrals using different methods allows us to check our work and ensure the accuracy of our results. It also helps us develop a deeper understanding of the concepts and techniques involved in integration.

4. Can completing the square be used for all integrals?

No, completing the square is only applicable to integrals that involve a quadratic term. For other types of integrals, different methods such as substitution or integration by parts may be more appropriate.

5. How do you know when to use completing the square in an integral?

Completing the square is most commonly used when the integrand contains a quadratic term and the coefficient of the squared term is not a perfect square. It is also useful when the integrand contains both a constant term and a linear term.

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