definite integral

May 2009
959
362
What's the trick to showing that \(\displaystyle \int^{2}_{0} \big( \sqrt{x^{3}+1} + \sqrt[3] {x^{2}+2x} \big) \ dx = 6 \) ? Is there a clever substitution?
 
Apr 2009
37
2
What's the trick to showing that \(\displaystyle \int^{2}_{0} \big( \sqrt{x^{3}+1} + \sqrt[3] {x^{2}+2x} \big) \ dx = 6 \) ? Is there a clever substitution?
Maybe you have to solve for x after the u substitution?

I'm trying to solve it now.
 
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simplependulum

MHF Hall of Honor
Jan 2009
715
427
tricky integral but dummy subsitution .


\(\displaystyle
\int^{2}_{0} \big( \sqrt{x^{3}+1} + \sqrt[3] {x^{2}+2x} \big)~ dx
\)

\(\displaystyle = \int^{2}_{0} \sqrt{x^{3}+1} ~dx + \int^{2}_{0} \sqrt[3] {(x+1)^2-1} ~ dx \)

Sub . \(\displaystyle (x+1)^2 - 1 = y^3 \) in the second integral .

\(\displaystyle 2(x+1)~dx = 3y^2~dy \)

Thus we have

\(\displaystyle \int^{2}_{0} \left( \sqrt{x^{3}+1} + \frac{3}{2} ~\frac{x^3}{\sqrt{x^3+1}} \right) ~dx\)

\(\displaystyle = \frac{1}{2}\int^{2}_{0} \frac{5x^3+2}{\sqrt{x^3+1}}~dx \)

\(\displaystyle = \frac{1}{2}\int^{2}_{0} \frac{5x^4+2x}{\sqrt{x^5+x^2}}~dx \)

\(\displaystyle = \left[ \sqrt{x^5+x^2} \right]_0^2\)

\(\displaystyle = \sqrt{36} - 0 \)

\(\displaystyle = 6 \)
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Where, exactly, did you get lost?

It should be clear that \(\displaystyle x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1\) and so \(\displaystyle \int_0^2 \sqrt{x^3+ 1}+ \sqrt[3]{x^2+ 2x} dx= \int_0^2 \sqrt{x^3+1}dx+ \int_0^2\sqrt[3]{(x+1)^2- 1}dx\).

Now, make that very clever substitution simplependulum gives: \(\displaystyle y^3= (x+1)^2- 1 \). Then \(\displaystyle 3y^2 dy= 2(x+1)dx\) and \(\displaystyle \sqrt[3]{(x+1)^2- 1}=\sqrt[3]{y^3}= y\).

Since \(\displaystyle y^3= (x+1)^2- 1\), \(\displaystyle (x+1)^2= y^3+ 1\) and \(\displaystyle x+ 1= \sqrt{y^3+ 1}\). That means that \(\displaystyle 3y^2dy= 2(x+1)dx= 2\sqrt{y^3+ 1} dx\) so that \(\displaystyle dx= \frac{3y^2}{2\sqrt{y^3+ 1}}dy= dx\).

Also, when x= 0, \(\displaystyle y^3= (1+1)^2- 1= 0\) and when x= 2, \(\displaystyle y^3= (1+2)^2- 1= 8\) so that y= 0 and 2 are still the limits of integration.

Putting that all together, \(\displaystyle \int_0^2 \sqrt[3]{(x+1)^2- 1}dx\)\(\displaystyle = \int_0^2\left(\sqrt[3]{y^3+ 1}\right)\left(\frac{3y^2}{2\sqrt{y^3+ 1}}dy\right)\)\(\displaystyle \frac{3}{2}\int_0^2\frac{y^3}{\sqrt{y^3+ 1}}dy\).

Now, that "y" is a dummy index- the actual integral is a number, not a function of y so we are free to replace it with anything we want. If just replace "y" with "x" (I think that's the "dummy substitution" simplependulum refers to. His first substitution is not "dummy" in any sense!) we get
\(\displaystyle \int_0^2 \frac{x^3}{\sqrt{x^3+ 1}}dx\).

Now put those two integrals back together:
\(\displaystyle \int_0^2 \sqrt{x^3+ 1}+ \frac{3}{2}\frac{x^3}{\sqrt{x^3+ 1}}dx\)

To "add fractions" get a "common denominator" by multiplying both numerator and denominator by \(\displaystyle 2\sqrt{x^3+ 1}\): \(\displaystyle \sqrt{x^3+1}\frac{2\sqrt{x^3+ 1}}{2\sqrt{x^3+ 1}}\)\(\displaystyle = \frac{2x^3+ 2}{2\sqrt{x^3+ 1}}\). (In effect, we are "rationalizing" the numerator.)

That gives \(\displaystyle \int_0^2 \frac{2x^3+ 2}{2\sqrt{x^3+ 1}}+ \frac{3x^3}{2\sqrt{x^3+ 1}}dx\)\(\displaystyle = \int_0^2 \frac{5x^3+ 2}{2\sqrt{x^3+ 1}}dx\).

simplependulum's next step was to multiply both numerator and denominator by x: \(\displaystyle \int_0^2 \frac{x(5x^3+ 2)}{2x\sqrt{x^3+ 1}}dx\). Of course, inside the square root that x becomes \(\displaystyle x^2\) so we have
\(\displaystyle \int_0^2 \frac{5x^4+ 2x}{2\sqrt{x^5+ x^2}}dx\).

And simplependulum did that because then the derivative of \(\displaystyle x^5+ x^2\) is precisely \(\displaystyle 5x^4+ 2x\)!

The substitution \(\displaystyle u= x^5+ x^2\) gives \(\displaystyle du= (5x^4+ 2x)dx\) so that \(\displaystyle \int_0^2\frac{5x^4+ 2x}{2\sqrt{x^5+ x^2}}dx\)\(\displaystyle = \int_0^36 \frac{du}{2\sqrt{u}}= \frac{1}{2}\int_0^36 u^{-1/2}du\).

That's my explanation. I don't pretend that I would have found that myself!
 
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Apr 2009
37
2
Where, exactly, did you get lost?

It should be clear that \(\displaystyle x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1\) and so \(\displaystyle \int_0^2 \sqrt{x^3+ 1}+ \sqrt[3]{x^2+ 2x} dx= \int_0^2 \sqrt{x^3+1}dx+ \int_0^2\sqrt[3]{(x+1)^2- 1}dx\).

Now, make that very clever substitution simplependulum gives: \(\displaystyle y^3= (x+1)^2- 1 \). Then \(\displaystyle 3y^2 dy= 2(x+1)dx\) and \(\displaystyle \sqrt[3]{(x+1)^2- 1}=\sqrt[3]{y^3}= y\).

Since \(\displaystyle y^3= (x+1)^2- 1\), \(\displaystyle (x+1)^2= y^3+ 1\) and \(\displaystyle x+ 1= \sqrt{y^3+ 1}\). That means that \(\displaystyle 3y^2dy= 2(x+1)dx= 2\sqrt{y^3+ 1} dx\) so that \(\displaystyle dx= \frac{3y^2}{2\sqrt{y^3+ 1}dy= dx\).

Also, when x= 0, \(\displaystyle y^3= (1+1)^2- 1= 0\) and when x= 2, \(\displaystyle y^3= (1+2)^2- 1= 8\) so that y= 0 and 2 are still the limits of integration.

Putting that all together, \(\displaystyle \int_0^2 \sqrt[3]{(x+1)^2- 1}dx\)\(\displaystyle = \int_0^2\left(\sqrt[3]{y^3+ 1}\right)\left(\frac{3y^2}{2\sqrt{y^3+ 1}}dy\right)\)\(\displaystyle \frac{3}{2}\int_0^2\frac{y^3}{\sqrt{y^3+ 1}}dy\).

Now, that "y" is a dummy index- the actual integral is a number, not a function of y so we are free to replace it with anything we want. If just replace "y" with "x" (I think that's the "dummy substitution" simplependulum refers to. His first substitution is not "dummy" in any sense!) we get
\(\displaystyle \int_0^2 \frac{x^3}{\sqrt{x^3+ 1}}dx\).

Now put those two integrals back together:
\(\displaystyle \int_0^2 \sqrt{x^3+ 1}+ \frac{3}{2}\frac{x^3}{\sqrt{x^3+ 1}}dx\)

To "add fractions" get a "common denominator" by multiplying both numerator and denominator by \(\displaystyle 2\sqrt{x^3+ 1}\): \(\displaystyle \sqrt{x^3+1}\frac{2\sqrt{x^3+ 1}}{2\sqrt{x^3+ 1}}\)\(\displaystyle = \frac{2x^3+ 2}{2\sqrt{x^3+ 1}}\). (In effect, we are "rationalizing" the numerator.)

That gives \(\displaystyle \int_0^2 \frac{2x^3+ 2}{2\sqrt{x^3+ 1}}+ \frac{3x^3}{2\sqrt{x^3+ 1}}dx\)\(\displaystyle = \int_0^2 \frac{5x^3+ 2}{2\sqrt{x^3+ 1}}dx\).

simplependulum's next step was to multiply both numerator and denominator by x: \(\displaystyle \int_0^2 \frac{x(5x^3+ 2)}{2x\sqrt{x^3+ 1}}dx\). Of course, inside the square root that x becomes \(\displaystyle x^2\) so we have
\(\displaystyle \int_0^2 \frac{5x^4+ 2x}{2\sqrt{x^5+ x^2}}dx\).

And simplependulum did that because then the derivative of \(\displaystyle x^5+ x^2\) is precisely \(\displaystyle 5x^4+ 2x\)!

The substitution \(\displaystyle u= x^5+ x^2\) gives \(\displaystyle du= (5x^4+ 2x)dx\) so that \(\displaystyle \int_0^2\frac{5x^4+ 2x}{2\sqrt{x^5+ x^2}}dx\)\(\displaystyle = \int_0^36 \frac{du}{2\sqrt{u}}= \frac{1}{2}\int_0^36 u^{-1/2}du\).

That's my explanation. I don't pretend that I would have found that myself!
I got the part up until the substitution, I've never done one like that.
Makes more sense now haha, would have never figured that out.
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
Generalization :


\(\displaystyle \int_a^b [ f(x) + f^{-1}(x+k)]~dx \)


If we have \(\displaystyle f^{-1}(a+k)=a ~,~ f^{-1}(b+k) = b \) or equivalently

\(\displaystyle f(a) = a+k ~,~ f(b) = b+k \) for some constants \(\displaystyle k\) ,

then the integral is equal to :

\(\displaystyle bf(b) - af(a) = b(b+k)-a(a+k) = (b-a)(a+b+k) \)


Proof :

Sub. \(\displaystyle u = f^{-1}(x+k) \) in the second integral ,

\(\displaystyle f(u) = x+ k \)

\(\displaystyle f'(u)du = dx \)

From the condition we know the limits of integration remain unchanged , then it is equal to

\(\displaystyle \int_a^b ( f(x) + xf'(x) )~dx \)

\(\displaystyle = \left[ xf(x) \right]_a^b = bf(b)-af(a) \)

From \(\displaystyle f(a) = a+k ~,~ f(b) = b+k \) we obtain

\(\displaystyle (b-a)(a+b+k) \)

For example ,

\(\displaystyle \int_1^2 [ x\Gamma(x) + \Gamma^{-1}(x) ]~dx = 4\)