Question

can some one please help me do this
∫4x^2(2x^3-7)^2 dx
step by step?

Answer 1

Im supposed to find the indefinite integral

Answer 2

it can be rewritten

\[4\int\limits x^2(2x^3 -7)^2 dx\]

use substitution

let

\[u = 2x^3 - 7\]

then \[\frac{du}{dx} = 6x^2\]

so \[du = 6x^2 dx\]

or

\[x^2 dx = \frac{1}{6} du\]

so using the substitution

\[4timesfrac{1}{6} \int\limits u^2 du\]

Answer 3

oops last line should read\[4 \times \frac{1}{6} \int\limits u^2 du\]

Answer 4

thats not the answer in my book....

Answer 5

which gives

\[\frac{2}{3} \times \frac{1}{3} u^3 + c\]

and \[u = 2x^3 - 7 \]

gives

\[\frac{2}{9} (2x^3 - 7)^3 + C\]

Answer 6

hope thats the answer

Answer 7

haha yeah it is.

Answer 8

so its all about the correct substitution... and working from there

Answer 9

to check your solution you can differentiate

Answer 10

ummm can you please explain how to get to step 2? haha sorry i'm retarded.

Answer 11

is that the derivative..?

Answer 12

no after you substitute u.

Answer 13

is it

\[4 \times \frac{1}{6} \int\limits u^2 du\]

Answer 14

perhaps if you draw it it may be easier

Answer 15

no the
\[du/dx=6x^2\]
part

Answer 16

how did you get that?

Answer 17

ok.... look at the integral in the rewritten form

\[4 \int\limits x^2(2x^3 - 7) dx\]

I can group things

\[4 \int\limits (2x^3 - 7) x^2dx\]

and if \[u = 2x^3 - 7\]

I need to find the derivative of u and get it to match x^2 dx in the integral

Answer 18

so

\[\frac{du}{dx} = 6x^2\]

manipulating the derivative

\[du = 6x^2 dx\]

but I only want x^2 dx... so I need to divide both sides by 6

hence

\[\frac{1}{6} du = x^2 dx\]

so I really make 2 substitutions 1 for \[u = 2x^3 - 7\]

and one for

\[\frac{1}{6} du = x^2 dx\]

Answer 19

HOW DID YOU DO THAT FIRST PART?

Answer 20

if it was \[y = 2x^3 - 7\]

can you differentiate it with respect to y...?

Answer 21

NO CAN YOU HELP ME DO THAT

Answer 22

ok.... then is is a big issue.... to be able to do integral calculus you need to know the opposite.. differential calculus... and Its a huge topic...

for this question the theory is

\[y = x^n\]

then \[\frac{dy}{dx} = nx^{n-1}\]

Answer 23

4∫x2(2x3−7)2dx
4∫x2(4x^6-28x^3+49)dx
4∫4x^8-28x^5+49x^2)dx
4[4∫x^8dx-28∫x^5dx+49∫x^2dx]

∫x^n=x^n+1/n+1

4[4(x^9/9-28x^6/6+49x^3/3]
16x^9/9-4*28x^6/6+4*49x^3/3

Answer 24

Caitlin1204 understood