Homework Help (Integration by Substitution)

L

Lightfall

New member
Joined
Sep 6, 2010
Messages
4
Hello, I am new to the forums. I am currently a college freshman taking calculus II and my first homework assignment's penultimate problem has me stumped. It asks:

Evaluate the definite integral:

dx/(1+sqrt[x])^4

with 0 and 1 as lower and upper bounds, respectively.

I know the answer is 1/6, but I have absolutely no clue how to arrive at that conclusion. For example I substitute a variable (My book uses "u") for (1+sqrt[x]) or the entire quantity (1+sqrt[x])^4, I would have nothing to be the "du". Also, I reviewed my integration charts and cannot manipulate this integral into anything that resembles a formula I've learned. When I put this integral as an indefinite one into Wolfram Mathematica's online integrator, I got back a mess involving logarithms. By the same token, putting it as an indefinite into my TI-89 gives -(3x+1)/[3(sqrt[x]+1)]^3.

I have a feeling that the very fact that this problem is given under the definite integral section means that it involves some trickery rather than the old fashion "integrate as indefinite and change the limits, then apply the 2nd fundamental calc theorem". Still, I cannot figure out what the trick is.

Thank you all in advance for your assistance.
 
Lightfall said:
Hello, I am new to the forums. I am currently a college freshman taking calculus II and my first homework assignment's penultimate problem has me stumped. It asks:

Evaluate the definite integral:

dx/(1+sqrt[x])^4

with 0 and 1 as lower and upper bounds, respectively.

I know the answer is 1/6, but I have absolutely no clue how to arrive at that conclusion. For example I substitute a variable (My book uses "u") for (1+sqrt[x]) or the entire quantity (1+sqrt[x])^4, I would have nothing to be the "du". Also, I reviewed my integration charts and cannot manipulate this integral into anything that resembles a formula I've learned. When I put this integral as an indefinite one into Wolfram Mathematica's online integrator, I got back a mess involving logarithms. By the same token, putting it as an indefinite into my TI-89 gives -(3x+1)/[3(sqrt[x]+1)]^3.

I have a feeling that the very fact that this problem is given under the definite integral section means that it involves some trickery rather than the old fashion "integrate as indefinite and change the limits, then apply the 2nd fundamental calc theorem". Still, I cannot figure out what the trick is.

Thank you all in advance for your assistance.
Click to expand...

\(\displaystyle \int_0^1\frac{dx}{(1+\sqrt{x})^4}\)

substitute

\(\displaystyle tan^2(\theta) = \sqrt{x}\)

\(\displaystyle x = tan^4(\theta)\)

and continue.....
 
Is there an alternative method to solving this? Subbing trig terms doesn't come until quite a few sections down the road in my book.
 
Lightfall said:
Is there an alternative method to solving this? Subbing trig terms doesn't come until quite a few sections down the road in my book.
Click to expand...

You are in CalcII - and you have not covered trig substitution in CalcI!!!!!
 
Correction, I STARTED calc 2, quite literally since this is the first day. And no, I never covered trig subs in high school when I took calc I, so bear with me here.
 
\(\displaystyle \int \frac{dx}{(1 + \sqrt{x})^4} \ = \ 2\int \frac{udu}{(1 +u)^4} \ = \ \int\frac{(1+u -1)}{(1+u)^4} du \ = \ \int\frac{1}{(1+u)^3}du \ - \ \int\frac{1}{(1+u)^4} du\)

Now continue....
 
\(\displaystyle \int_{0}^{1}\frac{dx}{(1+x^{1/2})^4}\)

\(\displaystyle Let \ u \ = \ 1 \ + \ x^{1/2}, \ then \ du \ = \ \frac{dx}{2x^{1/2}} \ = \ \frac{dx}{2(u-1)}, \ x^{1/2} \ = \ u-1.\)

\(\displaystyle Hence, \ we \ have \ \int_{1}^{2}\frac{2(u-1)du}{u^4} \ = \ 2\int_{1}^{2}\frac{(u-1)du}{u^4} \ = \ 2\int_{1}^{2}u^{-4}(u-1)du\)

\(\displaystyle = \ 2\int_{1}^{2}[u^{-3}-u^{-4}]du \ = \ 2\bigg[\frac{1}{3u^3}-\frac{1}{2u^2}\bigg]_{1}^{2} \ = \ 2\bigg[\frac{1}{24}-\frac{1}{8}-\bigg(\frac{1}{3}-\frac{1}{2}\bigg)\bigg] \ = \ \frac{1}{6}, \ QED.\)
 
You must log in or register to reply here.
Top