Integration using substitution (1 Viewer)

cloud edwards · Jun 17, 2017

Question: (refer to attachment), Evaluate the intergral using u=x^4 -3.

I did this question without subbing in the x=0 & x=1 and got a fairly reasonable answer being: 211/20. I also checked with a couple of online integration calculators.

HOWEVER,
I realized i was supposed to sub in the x values into u=x^4-3 to get the boundaries. But when i did this, i got a ridiculous number being: -174,000 something.

PLEASE HELP!!!

InteGrand · Jun 17, 2017

cloud edwards said:
Question: (refer to attachment), Evaluate the intergral using u=x^4 -3.

I did this question without subbing in the x=0 & x=1 and got a fairly reasonable answer being: 211/20. I also checked with a couple of online integration calculators.

HOWEVER,
I realized i was supposed to sub in the x values into u=x^4-3 to get the boundaries. But when i did this, i got a ridiculous number being: -174,000 something.

PLEASE HELP!!!

$\noindent Let $I = \int_{0}^{1}x^{3} \left(x^{4} - 3\right)^{4}\, \mathrm{d}x$. Let $u = x^{4} - 3$, so $\mathrm{d}u = 4x^{3}\, \mathrm{d}x \Rightarrow x^{3}\, \mathrm{d} x = \frac{1}{4}\mathrm{d}u $. \textsl{For the limits, when $x = 0$, $u = 0 - 3 = -3$, and when $x = 1$, $u = 1 - 3 = -2$.} So after the substitution, the integral becomes:$

$I = \int_{-3}^{-2}\frac{1}{4} u^{4}\,\mathrm{d}u.$

$\noindent Evaluating this gives us an answer of $\frac{1}{4}\times \frac{1}{5}\left(3^{5} - 2^{5}\right) = \frac{211}{20}$.$

Drongoski · Jun 17, 2017

let u = x^4 - 3 when x=0 u = -3 when x = 1 u=-2

So in your substituted work, you have to change the limits of integration to -3 and -2 resp.

Or you can do it my way without substitution:

$\int ^1 _ 0 x^3(x^4-3)^4dx = \frac{1}{4} \int ^1 _0 (x^4-3)^4d(x^4-3) = \frac {1}{4 \times 5} \left [(x^4 - 3)^5 \right ]^1 _0 = \frac {211}{20}$

cloud edwards · Jun 17, 2017

Thanx guys!

Integration using substitution (1 Viewer)

cloud edwards

Member

Attachments

InteGrand

Well-Known Member

Drongoski

Well-Known Member

cloud edwards

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)