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]

Consider the definite integral ∫202x(4−x2)1/5 dx.

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.

But from what there is below it appears you mean $$\int_2^0 2x(4- x^2)^{1/5}dx$$. Is that right?

If you cannot use Latex, at least use the standard ASCII "_" for subscripts and "^" for superscripts:
integral_2^0 2x(4- x^2)^(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

What in the world is "Preview Change entry mode"? Is this just copied arbitrarily from the computer screen?
Yes, if you make the substitution u= 4- x^2 them du= -2x dx

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)=

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 $$(4- x^2)^{1/5}(2xdx)= -(4- x^2)^{1/5}(-2xdx)$$.
Since $$4- x^2= u$$ and $$-2xdx= du$$, that is $$-u^{1/5}du$$.

Preview Change entry mode
Finally, use this work to compute

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