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

- jennychan12

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

- Stacey Warren - Expert brainly.com

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- jennychan12

I don't think you can use u-sub here. cuz then it'd be f(u) = u*sinu..

- jennychan12

and i haven't learned trig sub or integration by parts yet.

- anonymous

jenny you are a liar i think

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## More answers

- jennychan12

-_- i haven't. but i can do integration by parts. but my teacher never taught it.

- anonymous

becos that problem required fresnal integral

- jennychan12

what?

- anonymous

Use \(u = x^2\). Then \(x = \sqrt{u}\)

- anonymous

u can use u sub on that

- jennychan12

\[f(x) = x^2\sin(x^2) \]

- jennychan12

if u = x^2 then du = 2xdx

- anonymous

well what method are you allowed to use if you can't use trig or parts

- jennychan12

that's what i don't understand. u-sub doesn't really work...

- anonymous

so ur on u-sub section of the book?

- jennychan12

no it's a review packet. i learned area, u-sub, trapezoid rule, and simpson's rule

- anonymous

thats the way it is

- anonymous

not an exponent

- jennychan12

no it's \[f(x) = x^2\sin(x^2)\]

- anonymous

This isn't a candidate for u sub.

- anonymous

u = x^2 du = 2x dx cant use u du sub

- anonymous

no place to apply a substitution

- anonymous

\[ \frac{1}{2}\int \sqrt{u}\sin(u)du \]Isn't helpful.

- anonymous

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- jennychan12

-_- it was in the textbook in the chapter 5. we don't do trig sub/parts until chapter 8...

- anonymous

whats the name of the textbook

- jennychan12

although i know how to do parts but my teacher never taught it.

- jennychan12

stewart

- anonymous

7th edition?

- jennychan12

5th

- anonymous

which chapter section?

- jennychan12

chapter 5 review #49

- anonymous

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

- anonymous

fresnal function is introduced in section 5.3

- anonymous

lol wolframalpha? xD

- jennychan12

haha same 5.3 is fundemental theorum

- anonymous

go to 5.3 examlpe number 3

- jennychan12

we don't use stewart. we use larson. my teacher just photocopied this from the stewart textbook

- anonymous

thats illegail

- anonymous

but u gonna have to look up fresnal function elsewhere then

- jennychan12

she owns like ten stewart textbooks. -_- i'll just use parts then...

- anonymous

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

- anonymous

wait let me rewrite that

- anonymous

\[F(x) = \int\limits\limits_{1}^{x} t ^{2} \sin(t ^{2}) dt\]

- anonymous

That way when it is evaluated at F(1), you still get 0. And the derivative is the original function.

- anonymous

If you used Simpson's rule or Midpoint rule on that, you'd get a nasty summation but it is within the rules.

- jennychan12

oh wow. thanks @LogicalApple that's the answer in the book -___-

- anonymous

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.

- jennychan12

yes but i never thought of it that way... *sigh*

- anonymous

Stewart doesn't mess around.. I'm in chapter 5 on the 7th edition.

- jennychan12

oh same here i think. my teacher uses both stewart and larson. we teach from larson.

- anonymous

I will have to check that one out. Anyway good luck with your studies!

- jennychan12

thanks :)

- anonymous

I think it's sort of an unfair question. Should have specified a bit more the rigor.

- jennychan12

same. my teacher likes stewart because they have "more pictures" and "good problems". which means that she puts them on quizzes and tests. :[

- jennychan12

i guess it just makes u think about the concept more?

- anonymous

If they said something like "in terms of an indefinite integral" then it would have been fair.

- anonymous

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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