Integration of Quadratic Expressions |
Intro
An irreducible quadratic has no factors.
When the integrand includes an irreducible quadratic in the denominator, we
complete the square and use substitution. Since completing the square turns
the irreducible quadratic into a sum or difference of squares, we sometimes use
trig substitutions to solve the integral especially when the quadratic is square rooted.
.
Example 1
Evaluate
When we complete the square on the quadratic denominator we get
x² – 6x + 9 – 9 + 13 = (x – 3)² + 4, so now the integral is
so we use a substitution
Let u = x – 3, so du = dx and x = u + 3
This last statement means that 2x – 1 = 2(u + 3) – 1 = 2u + 5
Now our integral is
Using the substitution w = u² + 4, for the 1st integral, we get
ln (u² + 4) + (5/2) arcTan (u/2) and now we "unsubstitute"
The final answer is
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Example 2
Evaluate
To complete the square here, we first factor out a negative from the x terms, since
every perfect square includes a positive x².
So 8 + 2x – x^{ 2} = 8 – (x² – 2x) = 8 – (x² – 2x + 1) + 1 = 9 – (x – 1)².
Using the substitution u = x – 1 and du = dx, the integral becomes
and this integrates to .
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Example 3
Now we do one which requires a trig substitution to solve.
Evaluate
When we complete the square on the trinomial we get (x + 4)² – 16 + 25 = (x + 4)² + 9.
Using the substitution u = x + 4 and du = dx, the integral becomes
Now, with the trig sub u = 3 tan A and du = 3 sec² A dA , we get
and since 1 + tan² A = sec² A
we get
To get back to x's, the original variable we'll use the triangle created by the
substitution u = 3 tan A or u/3 = tan A. We'll also replace u with x + 4.
The answer then is
Practice
Evaluate these integrals by completing the square on the irreducible quadratic.
1) | 2) |
3) | 4) |
5) Find the area bounded by the graphs of , y = 0, x = 0, and x = 1. hint: use the same technique as for #4. | |
6) The velocity at time t of a point moving along a coordinate line is How far does the point travel during the interval t = 0 to t = 5 seconds? |
Solutions
1) x² – 4x + 8 = (x – 2)² + 4. so the integral I = I = ½ arcTan [½(x – 2)] + C |
2) 7 + 6x – x² = 16 – (x – 3)² so the integral I = I = arcSin [¼(x – 3)] + C |
3) 9 – 8x – x² = 25 – (x + 4)² so I = When we break this up into 2 integrals, I = I = |
4) x³ – 1 = (x – 1)(x² + x + 1), we complete the square on the quadratic x² + x + 1. x² + x + 1 = (x + ½)² + ¾, now we set u = x + ½ and du = dx. This makes (x – 1) = (u – 3/2), so I =. Now we use partial fraction decomposition to solve the integrals. I = |
5) Using the technique of #4, we get
Note we changed the limits of integration to values of u. square units. | |
. 6) |
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