Question

Differentiation of Integration

Answer 1

@freckles

Answer 2

Compute the derivative. \[\frac{ d }{ dt }\int\limits_{t ^{2}}^{t ^{3}}\sin(x ^{2})dx\]

Answer 3

wouldnt t be just 0?

Answer 4

k
\[\\ \text{ Let } F'=f \\ \int\limits_{a(x)}^{b(x)} f(x) dx= F(x)|_{a(x)}^{b(x)}=F(b(x))-F(a(x)) \\ \text{ now differentiating } \int\limits_{a(x)}^{b(x)}f(x) dx \text{ is the same as differentiating } \\ F(b(x))-F(a(x)) \\ \text{ since they are the same expression } \\ \frac{d}{dx}(F(b(x))-F(a(x)) =\frac{d}{dx}F(b(x))-\frac{d}{dx}F(a(x)) \\ \text{ we need the chain rule for both terms } \\ b'(x) \cdot F'(b(x))-a'(x)\cdot F'(a(x)) \\ \\ \text{ recall } F'=f \\ \frac{d}{dx} \int\limits_{a(x)}^{b(x)}=b'(x) f(b(x))-a'(x) \cdot f(a(x))\]

Answer 5

oops
\[\frac{d}{dx} \int\limits_{a(x)}^{b(x)} f(t) dt=b'(x) f(b(x))-a'(x) f(a(x))\]

Answer 6

Want me to give you the options?

Answer 7

also note I should I wrote
\[\int\limits_{a(x)}^{b(x)} f(t) dt \text{ earlier }\]
you aren't suppose to have the same variable in the limits as you have for what you are integrating withe respect to
but yea everything else is the same

Answer 8

you don't have to

Answer 9

you can apply what I said to find the answer though

Answer 10

it is just the fundamental theorem of calculus

Answer 11

I have basically written a formula above for you to use
but I hope you understand how the formula was derived above

Answer 12

so i just use this here b′(x)f(b(x))−a′(x)⋅f(a(x)) to find the answer?

Answer 13

and that is because of the first fundemental theorem

Answer 14

well yours will be in terms of t instead of x but yes
\[\frac{d}{dx} \int\limits_{a(x)}^{b(x)} f(t) dt = b'(x) f(b(x))-a'(x)f(a(x))\]

Answer 15

okay give me a minute

Answer 16

also I don't know which part your calculus book puts first

Answer 17

but I would say it is by both parts of the fundamental theorem of calculus

Answer 18

okay soit looks like ittl be this one \[3t ^{2}sint ^{3}-2tsint ^{4}\]

Answer 19

ok your insides are off
but your did good on differentiating your limits

Answer 20

would it be possible that the 3t^2 would be negative?

Answer 21

the inside is x^2
your upper limit is t^3
if we plug t^3 where x is in x^2 we have \[(t^3)^2=t^{3 \cdot 2} \text{ by law of exponents }\]

Answer 22

your second inside is actually fine
it looks like you did 2*2 there ..

Answer 23

wait the only other option that looks close uses cos

Answer 24

have you fixed the inside yet?

Answer 25

so does it use cos because of the derivative of sin is cos

Answer 26

\[\frac{ d }{ dt }\int\limits\limits_{t ^{2}}^{t ^{3}}\sin(x ^{2})dx \\ =3t^2 \sin((t^3)^2)-2t \sin((t^2)^2) \\ =3t^2 \sin(t^6)-2t \sin(t^4)\]

Answer 27

no cosine

Answer 28

isnt that what i put earlier?

Answer 29

you had t^3 in the first inside

Answer 30

instead of t^6

Answer 31

ohhhhh okay my bad yah i meant to put the 6

Answer 32

Thank you

Answer 33

so you are not finding that this matches one of the answers?
we could use sin is odd
\[=3t^2 \sin(t^6)+2t \sin(-t^4)\]

we could factor ...
there are a few things we can do

Answer 34

yah it is

Answer 35

the first one

Answer 36

oh okay

Answer 37

haha thanks

Answer 38

np

Answer 39

I have three questions left if my stupidity hasn't annoyed you enough already.

Answer 40

I can help with one more
but then I have to eat

Answer 41

okay thanks my professor makes us do our study guides, but doesnt give us credit

Answer 42

lol so that means you don't have to do the study guides? but it is probably worth your while to do them so you can pass the test or at least higher your chances of passing the test