Integration Question (1 Viewer)

[joesmith1975]

[Integration symbol]high 1 low 0 of x+1/3(x^2 + 2x -1)^2 dx , use u =x^2 +2x -1



Answer is -1/4

But im geting 1/12

Please help!

 

[Riviet]

Don't forget that when using a substitution of u to find a definite integral, you need to also change the limits to that of u.
ie for u=x²+2x-1
when x=0, u=-1 and when x=1, u=2
So you should get this line in the process:
2
/
| du/(6u²)
/
-1

Is that what you had?

Edit: sorry, I had to correct myself a few times.

 

[joesmith1975]

no for that i got

1 / 2 du
- | ___
2 /-1 3u^2




But i think thats the same but im geting 1/12 as answer

 

[Riviet]

Alright I got it:

1/6 x integral (lower bound -1 upper 2) du/u² = 1/6.[-1/u] upper 2, lower -1
= 1/6.[(-1/2)-(1)]
= -1/4, watch the negative signs when substituting -1 in.
Whoah it's that late already, I'm off now, hope that helped.

 

[joesmith1975]

thanks for that i didnt see one of the negative signs there is 3{for the -1 value} of them still making it -1

Thanks alot

 

[pLuvia]

[Integration symbol]high 1 low 0 of x+1/3(x^2 + 2x -1)^2 dx , use u =x^2 +2x -1



Answer is -1/4

But im geting 1/12

Please help!

1
∫ (x+1)/3(x²+2x-1)² dx
0

Let u = x²+2x-1 du=2x+2 dx

1
1/6 ∫ du/u[/sup]2[/sup]
0

= 1/6 [ -1/u ]¹₀
= 1/6 [ -1/(x²+2x-1)]¹₀

Subing it in

= 1/6 [ -1/2 - 1]
= -1/4