- #1
okkvlt
- 53
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suppose i want to find the following integral:
7
[tex]\int[/tex]x dx
3
now suppose for some demented reason i decided not to do it straightforward and get (49-9)/2=20
instead i use the substitution x=u2+4u+5
giving
u1
[tex]\int[/tex](u2+4u+5)(2u+4)du
u0
u1
[tex]\int[/tex]2u3+12u2+26u+20 du
u0
the indefinite integral is .5u4+4u3+13u2+20u
now its time to find the limits of integration,
by the quadratic formula,
u1=(-4[tex]\pm[/tex][tex]\sqrt{24}[/tex])/2=0.44948974278318,-4.44948974278318
u0=(-4[tex]\pm[/tex][tex]\sqrt{8}[/tex])/2=-0.5857864376269, -3.4142135623731
what happens when you enter these limits of integration?
Here is what's fascinating:
if i use both values of u1 as the limits of integration, i get 0 (the case is the same with using both values of u0).
if i use either value of u1 as the upper limit and either value of u0 as the lower limit, i get the correct answer of 20. there is a total of 4 ways to correctly get the answer. it doesn't matter which of the values i use as long as i use a value of u1 as the upper limit and a value of u0 as the lower limit.
you can go ahead and test it to see what i mean.
7
[tex]\int[/tex]x dx
3
now suppose for some demented reason i decided not to do it straightforward and get (49-9)/2=20
instead i use the substitution x=u2+4u+5
giving
u1
[tex]\int[/tex](u2+4u+5)(2u+4)du
u0
u1
[tex]\int[/tex]2u3+12u2+26u+20 du
u0
the indefinite integral is .5u4+4u3+13u2+20u
now its time to find the limits of integration,
by the quadratic formula,
u1=(-4[tex]\pm[/tex][tex]\sqrt{24}[/tex])/2=0.44948974278318,-4.44948974278318
u0=(-4[tex]\pm[/tex][tex]\sqrt{8}[/tex])/2=-0.5857864376269, -3.4142135623731
what happens when you enter these limits of integration?
Here is what's fascinating:
if i use both values of u1 as the limits of integration, i get 0 (the case is the same with using both values of u0).
if i use either value of u1 as the upper limit and either value of u0 as the lower limit, i get the correct answer of 20. there is a total of 4 ways to correctly get the answer. it doesn't matter which of the values i use as long as i use a value of u1 as the upper limit and a value of u0 as the lower limit.
you can go ahead and test it to see what i mean.
Last edited: