- #1

okkvlt

- 53

- 0

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=u

giving

u

[tex]\int[/tex](u

u

u

[tex]\int[/tex]2u

u

the indefinite integral is .5u

now its time to find the limits of integration,

by the quadratic formula,

u

u

what happens when you enter these limits of integration?

Here is what's fascinating:

if i use both values of u

if i use either value of u

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=u

^{2}+4u+5giving

u

_{1}[tex]\int[/tex](u

^{2}+4u+5)(2u+4)duu

_{0}u

_{1}[tex]\int[/tex]2u

^{3}+12u^{2}+26u+20 duu

_{0}the indefinite integral is .5u

^{4}+4u^{3}+13u^{2}+20unow its time to find the limits of integration,

by the quadratic formula,

u

_{1}=(-4[tex]\pm[/tex][tex]\sqrt{24}[/tex])/2=0.44948974278318,-4.44948974278318u

_{0}=(-4[tex]\pm[/tex][tex]\sqrt{8}[/tex])/2=-0.5857864376269, -3.4142135623731what happens when you enter these limits of integration?

Here is what's fascinating:

if i use both values of u

_{1}as the limits of integration, i get 0 (the case is the same with using both values of u_{0}).if i use either value of u

_{1}as the upper limit and either value of u_{0}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: