Calcule \iint_{R} f(x, y) d x d y para:
f(x, y)=x y \cos \left(x^{2}\right), R=\left\{(x, y) \in \mathbb{R}^{2} \mid 0 \leqslant x \leqslant 1, x^{2} \leqslant y \leqslant 1\right\} ;

Passo 1

O enunciado pede
\iint_{R}f(x,y)\,dx\,dy
para f(x,y)=x\,y\cos\bigl(x^2\bigr) e
R=\bigl\{(x,y)\mid 0\le x\le1,\;x^2\le y\le1\bigr\}.
Vamos usar a integral iterada em y primeiro e depois em x.

Passo 2

Escrevendo a integral iterada, temos
\iint_{R}f(x,y)\,dx\,dy =\int_{x=0}^{1}\!\Bigl[\int_{y=x^2}^{1}x\,y\cos\bigl(x^2\bigr)\,dy\Bigr]dx.

Passo 3

Na integral interna y\mapsto x\,\cos(x^2) é constante, então
\int_{y=x^2}^{1}x\,y\cos\bigl(x^2\bigr)\,dy =x\cos\bigl(x^2\bigr)\int_{x^2}^{1}y\,dy =x\cos\bigl(x^2\bigr)\,\frac{1 - x^4}{2}.
Logo
\iint_{R}f =\int_{0}^{1}x\cos\bigl(x^2\bigr)\,\frac{1 - x^4}{2}\,dx =\frac12\int_{0}^{1}x\bigl(1 - x^4\bigr)\cos\bigl(x^2\bigr)\,dx.

Passo 4

Para a integral em x, faça t=x^2, daí dt=2x\,dx e x\,dx=dt/2. Quando x=0 vem t=0 e quando x=1 vem t=1. Assim
\frac12\int_{0}^{1}x\bigl(1 - x^4\bigr)\cos\bigl(x^2\bigr)\,dx =\frac14\int_{0}^{1}\bigl(1 - t^2\bigr)\cos t\,dt.

Passo 5

Calcule
\int_{0}^{1}\bigl(1 - t^2\bigr)\cos t\,dt =\int_{0}^{1}\cos t\,dt -\int_{0}^{1}t^2\cos t\,dt.
Temos
\int_{0}^{1}\cos t\,dt =\bigl[\sin t\bigr]_{0}^{1} =\sin1,
e por partes,
\int t^2\cos t\,dt =t^2\sin t-2\int t\sin t\,dt =t^2\sin t+2t\cos t-2\sin t.
Avaliando de 0 a 1, dá
\int_{0}^{1}t^2\cos t\,dt =\bigl[\;t^2\sin t+2t\cos t-2\sin t\;\bigr]_{0}^{1} =2\cos1-\sin1.

Passo 6

Logo
\int_{0}^{1}\bigl(1 - t^2\bigr)\cos t\,dt =\sin1-\bigl(2\cos1-\sin1\bigr) =2\sin1-2\cos1 =2\bigl(\sin1-\cos1\bigr).
Multiplicando por \tfrac14 vem
\iint_{R}f(x,y)\,dx\,dy =\frac14\cdot2\bigl(\sin1-\cos1\bigr) =\frac{\sin1-\cos1}{2}.

Resposta

\frac{\sin1-\cos1}{2}.

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