What is the substitution for the definite integral ∫202x(4−x2)1/5 dx?

[Yousra1]

Consider the definite integral ∫202x(4−x2)1/5 dx.
What is the substitution to use? u= 4-x^2
Preview Change entry mode (There can be more than one valid substitution; give the one that is the most efficient.)
For this correct choice, du/dx= -2x
Preview Change entry mode
If we make this substitution, then the integral becomes of the form ∫baf(u)du. What are a, b and f(u)?
a= 4
b= 0
f(u)=
Preview Change entry mode
Finally, use this work to compute

∫202x(4−x2)1∕5 dx=
Preview Change entry mode
Give the exact value.

 

[HallsofIvy]

Re: calculus

What is this? Did you just post what appeared on a computer screen?

[tex]<blockquote data-attributes="member: 710624" data-quote="Yousra" data-source="post: 6751605" cite="https://www.physicsforums.com/goto/post?id=6751605" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Yousra said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Consider the definite integral ∫202x(4−x2)1/5 dx. </div> </div> </blockquote> What you have written is $$\int 202 x(4- 2x)(1/5)dx$$. That's not a "definite integral" because there are no limits of integration.<br /> <br /> But from what there is below it appears you mean $$\int_2^0 2x(4- x^2)^{1/5}dx$$. Is that right?<br /> <br /> If you cannot use Latex, at least use the standard ASCII "_" for subscripts and "^" for superscripts:<br /> integral_2^0 2x(4- x^2)^(1/5) dx<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> What is the substitution to use? u= 4-x^2<br /> Preview Change entry mode (There can be more than one valid substitution; give the one that is the most efficient.)<br /> For this correct choice, du/dx= -2x<br /> Preview Change entry mode </div> </div> </blockquote> What in the world is "Preview Change entry mode"? Is this just copied arbitrarily from the computer screen?<br /> Yes, if you make the substitution u= 4- x^2 them du= -2x dx<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> If we make this substitution, then the integral becomes of the form ∫baf(u)du. What are a, b and f(u)?<br /> a= 4<br /> b= 0<br /> f(u)= </div> </div> </blockquote> Yes, when x= 0, u= 4- 0^2= 4 and when x= 2, u= 4- 2^2= 0. The integrand appears to be $$2x(4- x^2)^{1/5}dx$$ which can be written $$(4- x^2)^{1/5}(2xdx)= -(4- x^2)^{1/5}(-2xdx)$$.<br /> Since $$4- x^2= u$$ and $$-2xdx= du$$, that is $$-u^{1/5}du$$.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Preview Change entry mode<br /> Finally, use this work to compute<br /> <br /> ∫202x(4−x2)1∕5 dx= <br /> Preview Change entry mode<br /> Give the exact value. </div> </div> </blockquote>[/tex]