Question

Calculus Help
Fan + Medal

Answer 1

u-sub

Answer 2

before that tho, you need algebra.
Factor out of 2 in 1st parenthesis.

Answer 3

\(\color{#000000}{ \displaystyle \int\limits_{-1}^0\left(2x^5+6x\right)^3\left(5x^4+3\right) dx }\)

Answer 4

have you covered u-substitution in your class?

Answer 5

Yes, we have

Answer 6

and I will assume you know algebra too, once you are in calculus class.

Answer 7

Can you factor this out of two?
\(\color{#000000}{ \displaystyle (2x^5+6x)^3 }\)

Answer 8

2x(x^4+3)^3(5x^4+3)

Answer 9

Or just 2?

Answer 10

yes, out of 2.
and that was not correct for 2x.

Answer 11

\(\color{#000000}{ \displaystyle \left(2x^5+6x\right)^3= \left(2[x^5+3x]\right)^3 =(2)^3\left(x^5+3x\right)^3=8\left(x^5+3x\right)^3 }\)

agree?

Answer 12

Yes.

Answer 13

this is because of the algebraic rule:
\(\color{#000000}{ \displaystyle \left(a\times b\right)^n=a^n\times b^n }\)

Answer 14

So, the integral will now become,
\(\color{#000000}{ \displaystyle 8\int\limits_{-1}^0\left(x^5+3x\right)^3\left(5x^4+3\right) dx }\)

Answer 15

In general, when you do a u-sub; suppose, you let: \(\color{#000000}{ \displaystyle u=f(x) }\)
then, \(\color{#000000}{ \displaystyle du=f'(x)~dx }\).

Correct?

Answer 16

Right.

Answer 17

Right, and here you can do a very convenient u-substitution.
\((\)Hint: What is the derivative of \(\color{#000000}{ \displaystyle x^5+3x }\) ? \()\)

Answer 18

5x^4+3

Answer 19

Yes, so what should your \(u\) be?

Answer 20

5x^4+3

Answer 21

No. Recall the setup in general.

if u = f(x)
then du = f'(x) dx

Answer 22

u= x^5+3x

Answer 23

Yes!

Answer 24

du=5x^4+3

Answer 25

you forgot the dx

Answer 26

I realized that after I sent it. du=5x^4+3 dx

Answer 27

Yes, and more precisely:
`  du = (5x^4+3) dx  `

(you are multiplying 5x^4 by dx as well)

Answer 28

Right

Answer 29

\(\color{#000000}{ \displaystyle 8\int\limits_{-1}^0\left(\color{blue}{x^5+3x}\right)^3\color{red}{\left(5x^4+3\right) dx} }\)

\(\color{#000000}{ \displaystyle u=x^5+3x }\)
\(\color{#000000}{ \displaystyle du=(5x^4+3) dx }\)

Answer 30

Now, find your new limits of integration.

Answer 31

If \(x=-1\), then (using the fact that \(u=x^5+3x\)), you get \(u=?\)

Answer 32

One second i was afk. im back

Answer 33

=-4

Answer 34

GOOD!

Then, find the upper limit for \(u\) (that corresponds to \(x=0\)). (Both of these you can achieve, by plugging the original limits of integ. into the expression for \(u\) (i.e. into x^5+3x).

Answer 35

that would equal 0

Answer 36

Yes!

Answer 37

So you get:
\(\color{#000000}{ \displaystyle 8\int\limits_{-4}^0\left(u\right)^3du }\)

Answer 38

the part that I previously colored with red (the second parenthesis and dx), goes away, replaced by du. And the parenthesis in ()^3 is replaced by u.

Answer 39

Now, integrate by power rule, and plug in these (new) limits of integration.

Answer 40

I have to leave right now. Sorry.