Question

Find the integral t(t^2+7)^2

Answer 1

you could expand it out; or just notice that your missing a 3 and a 2

Answer 2

I am confused by what you just said
the question is
\[\int\limits_{}t(t^2+7)^2\]

Answer 3

with dt after it

Answer 4

well, if we notice this as a x^2 type format; then we should see that it comes from something that looks like x^3/3

lets try that out and see what we get when we derive it

(t^2+7)^3
--------- derives to what?
3

Answer 5

t^2+7

Answer 6

not quite; bring down the exponent, subtract the 1 and pop out the derivative of the innards

Answer 7

(t^2+7)^2

Answer 8

good, and the chain rule pops out a derivative of the inside parts

Answer 9

otherwise:

t(t^2+7)^2 = t (t^4+14t^2+49)
= t^5 +14t^3 + 49t

\[\int t^5 +14t^3 + 49t\ dt\] is just as usable

Answer 10

most likely this is an example of a u-substitution .... which isnt needed but might help to claen things up

Answer 11

ya this chapter is over the substitution

Answer 12

say u = t^2 + 7

du/dt = 2t

du/2 = t dt

\[\int \frac{u^2}{2}du\]

Answer 13

in this format, its just a simple integration that you should be used to by now

Answer 14

This is the step I get lost at

Answer 15

the u subbing or the integration? :)

Answer 16

its a method that changes the way the integrand looks, while keeping the same value

Answer 17

replace all the t parts by its equivalent u parts

Answer 18

since we define u = t^2 + 7 ; wherever you see a t^2 + 7, replace it with a "u"

dt has to be replaced by du .... but an equivalent form of it

since u = t^2 + 7 , we can derive it and solve for dt

du/dt = 2t

du = 2t dt

dt/2t = du is good enough

Answer 19

When do we plug the t parts back in?

Answer 20

replace all that parts:\[\int t(t^2+7)^2dt\to \ \int t(u)^2\frac{du}{2t}\]

Answer 21

after we integrate up the u fillins, we replace the u parts by their t parts

Answer 22

Integrate the second expression you put? Correct?

Answer 23

itegrate either of them, but yes, the second one is spose to be the easier looking one

Answer 24

\[\int\limits t^5 +14t^3 + 49t\ dt=\frac {t^6}{6}+14\frac{t^4}{4}+49\frac{t^2}{2}+C\]

Answer 25

we should end up with:\[\frac{1}{6}u^3\]
and replaceing the u parts; we know u = t^2 + 7\[\frac{1}{6}(t^2+7)^3\]

Answer 26

ya thats what I got thanks!

Answer 27

dont forget the +C arbitrary constant at the end

Answer 28

\[\frac16 (t^2+7)^3=\frac {t^6}{ 6}+\frac{7t^4}{2}+\frac{49t^2}{2}+\frac{343}{6}+C\]

Answer 29

\[\frac {343} 6+C=C_1\]