Re: calculus
What is this? Did you just post what appeared on a computer screen?
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Yousra said:
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Consider the definite integral ∫202x(4−x2)1/5 dx.
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</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 />
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But from what there is below it appears you mean \int_2^0 2x(4- x^2)^{1/5}dx. Is that right?<br />
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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 />
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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
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</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 />
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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)=
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</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 />
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Preview Change entry mode<br />
Finally, use this work to compute<br />
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∫202x(4−x2)1∕5 dx= <br />
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Give the exact value.
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