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Find an antiderivative F of f(x) = x^2*sin(x^2) such that F(1) = 0.

Mathematics
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I don't think you can use u-sub here. cuz then it'd be f(u) = u*sinu..
and i haven't learned trig sub or integration by parts yet.
jenny you are a liar i think

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Other answers:

-_- i haven't. but i can do integration by parts. but my teacher never taught it.
becos that problem required fresnal integral
what?
Use \(u = x^2\). Then \(x = \sqrt{u}\)
u can use u sub on that
\[f(x) = x^2\sin(x^2) \]
if u = x^2 then du = 2xdx
well what method are you allowed to use if you can't use trig or parts
that's what i don't understand. u-sub doesn't really work...
so ur on u-sub section of the book?
no it's a review packet. i learned area, u-sub, trapezoid rule, and simpson's rule
thats the way it is
not an exponent
no it's \[f(x) = x^2\sin(x^2)\]
This isn't a candidate for u sub.
u = x^2 du = 2x dx cant use u du sub
no place to apply a substitution
\[ \frac{1}{2}\int \sqrt{u}\sin(u)du \]Isn't helpful.
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-_- it was in the textbook in the chapter 5. we don't do trig sub/parts until chapter 8...
whats the name of the textbook
although i know how to do parts but my teacher never taught it.
stewart
7th edition?
5th
which chapter section?
chapter 5 review #49
integral x^2 sin(x^2) dx = 1/4 (sqrt(2 pi) C(sqrt(2/pi) x)-2 x cos(x^2))+constant C(x) is the fresnal C integral Medal pls
fresnal function is introduced in section 5.3
lol wolframalpha? xD
haha same 5.3 is fundemental theorum
go to 5.3 examlpe number 3
we don't use stewart. we use larson. my teacher just photocopied this from the stewart textbook
thats illegail
but u gonna have to look up fresnal function elsewhere then
she owns like ten stewart textbooks. -_- i'll just use parts then...
I guess one way to approach it is: \[F(t) = \int\limits_{1}^{x} t ^{2} \sin(t ^{2})\] This way the integral evaluated at x = 1 results in 0, and its derivative is just the original function
wait let me rewrite that
\[F(x) = \int\limits\limits_{1}^{x} t ^{2} \sin(t ^{2}) dt\]
That way when it is evaluated at F(1), you still get 0. And the derivative is the original function.
If you used Simpson's rule or Midpoint rule on that, you'd get a nasty summation but it is within the rules.
oh wow. thanks @LogicalApple that's the answer in the book -___-
Well it makes sense, right? We don't actually have to take an integration. We know from the fundamental theorem of calculus (either part 1 or 2, i forget) that the derivative of F(x) here is just the inside of the integral evaluated at x. And for the limits of integration, we know that if we integrate from a to a we always get 0. So let a = 1, that way F(1) is an integral evaluated from 1 to 1.
yes but i never thought of it that way... *sigh*
Stewart doesn't mess around.. I'm in chapter 5 on the 7th edition.
oh same here i think. my teacher uses both stewart and larson. we teach from larson.
I will have to check that one out. Anyway good luck with your studies!
thanks :)
I think it's sort of an unfair question. Should have specified a bit more the rigor.
same. my teacher likes stewart because they have "more pictures" and "good problems". which means that she puts them on quizzes and tests. :[
i guess it just makes u think about the concept more?
If they said something like "in terms of an indefinite integral" then it would have been fair.
Yeah sometimes Stewart's questions are very very chapter-specific That's why I asked what chapter the problem was in to get an idea of the context of the question.

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