anonymous
  • anonymous
Evaluate exactly the value of the integral from negative 1 to 0 of the product of the cube of the quantity 2 times x to the 4th power plus 8 times x and 4 times x to the 3rd power plus 4, dx. Your work must include the use of substitution and the antiderivative.
Mathematics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
https://gyazo.com/a94388763f348d36040ce04fad93ecc0
anonymous
  • anonymous
@jim_thompson5910 u substitution with multiplication rule?
anonymous
  • anonymous
If so how would I approach it using u substitution?

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More answers

Zarkon
  • Zarkon
\[\Large u=2x^4+8x\]
anonymous
  • anonymous
@Zarkon du would be equal to 8x^3
Zarkon
  • Zarkon
no
anonymous
  • anonymous
oops wrote down +8 instead of 8x
anonymous
  • anonymous
So it would be 8x^3+8
anonymous
  • anonymous
So I need to multiply by 2?
Zarkon
  • Zarkon
almost
anonymous
  • anonymous
Then solve?
Zarkon
  • Zarkon
\[\Large du=(8x^3+8)dx\]
anonymous
  • anonymous
I usually forget to include that thanks
Zarkon
  • Zarkon
divide by 2
anonymous
  • anonymous
Multiply the part in the integral by 2
anonymous
  • anonymous
No?
Zarkon
  • Zarkon
\[\Large \frac{du}{2}=(4x^3+4)dx\] then substitute
anonymous
  • anonymous
Why?
anonymous
  • anonymous
\[2(4x^3 + 4) = du\]
Zarkon
  • Zarkon
\[2(4x^3 + 4)dx = du\] so \[(4x^3 + 4)dx = \frac{du}{2}\]
anonymous
  • anonymous
Why is dividing by 2 preferred?
Zarkon
  • Zarkon
because \((4x^3+4)dx\) is in the problem
Zarkon
  • Zarkon
you can now replace that part with du/2
anonymous
  • anonymous
I now have \[\int\limits_{-1}^{0}\frac{ u^3du}{ 2}\]
Zarkon
  • Zarkon
close
Zarkon
  • Zarkon
\[\Large \int\limits_{-1}^{0}(2x^4+8x)^3(8x^3+8)dx=\int\limits_{u(-1)}^{u(0)}u^3\frac{du}{2}\]
Zarkon
  • Zarkon
-1 and 0 are limits for x not u
anonymous
  • anonymous
What's the difference between what you did and I did?
anonymous
  • anonymous
My calc teacher didn't seem to do that step, kept it the same.
Zarkon
  • Zarkon
then he/she is wrong :)
Zarkon
  • Zarkon
you should double check your notes. I doubt that they kept them the same...unless they did a work around. But what you wrote is definitely wrong.
anonymous
  • anonymous
But we will be returning to x no?
Zarkon
  • Zarkon
only if you want to
Zarkon
  • Zarkon
you can write it in terms of just u or you can return back to x and use the original limits
anonymous
  • anonymous
Yes, that's the plan. How would I proceed with that?
Zarkon
  • Zarkon
find the antiderivative of \(\dfrac{u^3}{2}\)
anonymous
  • anonymous
\[\frac{ u^4 }{ 8 }\]
Zarkon
  • Zarkon
yes. Now replace u by what you set it equal to
anonymous
  • anonymous
I got -162 for final answer.
Zarkon
  • Zarkon
yep
anonymous
  • anonymous
Can I ask a quick question?
anonymous
  • anonymous
On something different?
Zarkon
  • Zarkon
what?
anonymous
  • anonymous
To find whether something concaves up or down you'd check the second derivative right?
Zarkon
  • Zarkon
if the second derivative exists...yes
anonymous
  • anonymous
Ok so what do I check it for? I have the second derivative but not sure how to check if it concaves up or down
Zarkon
  • Zarkon
it depends on where you evaluate it. it can be concave down on some parts and concave up on others
anonymous
  • anonymous
Should I make another post?
Zarkon
  • Zarkon
prob
anonymous
  • anonymous
Just made it and tagged you. Thanks

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