Question

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.

Answer 1

https://gyazo.com/a94388763f348d36040ce04fad93ecc0

Answer 2

@jim_thompson5910 u substitution with multiplication rule?

Answer 3

If so how would I approach it using u substitution?

Answer 4

\[\Large u=2x^4+8x\]

Answer 5

@Zarkon du would be equal to 8x^3

Answer 6

no

Answer 7

oops wrote down +8 instead of 8x

Answer 8

So it would be 8x^3+8

Answer 9

So I need to multiply by 2?

Answer 10

almost

Answer 11

Then solve?

Answer 12

\[\Large du=(8x^3+8)dx\]

Answer 13

I usually forget to include that thanks

Answer 14

divide by 2

Answer 15

Multiply the part in the integral by 2

Answer 16

No?

Answer 17

\[\Large \frac{du}{2}=(4x^3+4)dx\]

then substitute

Answer 18

Why?

Answer 19

\[2(4x^3 + 4) = du\]

Answer 20

\[2(4x^3 + 4)dx = du\]

so

\[(4x^3 + 4)dx = \frac{du}{2}\]

Answer 21

Why is dividing by 2 preferred?

Answer 22

because \((4x^3+4)dx\) is in the problem

Answer 23

you can now replace that part with du/2

Answer 24

I now have \[\int\limits_{-1}^{0}\frac{ u^3du}{ 2}\]

Answer 25

close

Answer 26

\[\Large \int\limits_{-1}^{0}(2x^4+8x)^3(8x^3+8)dx=\int\limits_{u(-1)}^{u(0)}u^3\frac{du}{2}\]

Answer 27

-1 and 0 are limits for x not u

Answer 28

What's the difference between what you did and I did?

Answer 29

My calc teacher didn't seem to do that step, kept it the same.

Answer 30

then he/she is wrong :)

Answer 31

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.

Answer 32

But we will be returning to x no?

Answer 33

only if you want to

Answer 34

you can write it in terms of just u or you can return back to x and use the original limits

Answer 35

Yes, that's the plan. How would I proceed with that?

Answer 36

find the antiderivative of \(\dfrac{u^3}{2}\)

Answer 37

\[\frac{ u^4 }{ 8 }\]

Answer 38

yes. Now replace u by what you set it equal to

Answer 39

I got -162 for final answer.

Answer 40

yep

Answer 41

Can I ask a quick question?

Answer 42

On something different?

Answer 43

what?

Answer 44

To find whether something concaves up or down you'd check the second derivative right?

Answer 45

if the second derivative exists...yes

Answer 46

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

Answer 47

it depends on where you evaluate it. it can be concave down on some parts and concave up on others

Answer 48

Should I make another post?

Answer 49

prob

Answer 50

Just made it and tagged you. Thanks