Examples from Fubini and iterated integrals All examples from Fubini and iterated integrals.
[ EXAMPLE ] Let
R = [ 0 , 1 ] × [ 0 , 2 ] , f ( x , y ) = x y 2 . R=[0,1]\times[0,2],
\qquad
f(x,y)=xy^2. R = [ 0 , 1 ] × [ 0 , 2 ] , f ( x , y ) = x y 2 . Integrating with respect to x x x
∬ R x y 2 d A = ∫ 0 2 ∫ 0 1 x y 2 d x d y = ∫ 0 2 [ x 2 y 2 2 ] 0 1 d y = ∫ 0 2 y 2 2 d y = 4 3 . \begin{aligned}
\iint_R xy^2\,dA
&=\int_0^2\int_0^1xy^2\,dx\,dy\\
&=\int_0^2\left[\frac{x^2y^2}{2}\right]_{0}^{1}dy\\
&=\int_0^2\frac{y^2}{2}\,dy
=\frac43.
\end{aligned} ∬ R x y 2 d A = ∫ 0 2 ∫ 0 1 x y 2 d x d y = ∫ 0 2 [ 2 x 2 y 2 ] 0 1 d y = ∫ 0 2 2 y 2 d y = 3 4 . Integrating with respect to y y y
∬ R x y 2 d A = ∫ 0 1 ∫ 0 2 x y 2 d y d x = ∫ 0 1 [ x y 3 3 ] 0 2 d x = ∫ 0 1 8 x 3 d x = 4 3 . \begin{aligned}
\iint_R xy^2\,dA
&=\int_0^1\int_0^2xy^2\,dy\,dx\\
&=\int_0^1\left[\frac{xy^3}{3}\right]_{0}^{2}dx\\
&=\int_0^1\frac{8x}{3}\,dx
=\frac43.
\end{aligned} ∬ R x y 2 d A = ∫ 0 1 ∫ 0 2 x y 2 d y d x = ∫ 0 1 [ 3 x y 3 ] 0 2 d x = ∫ 0 1 3 8 x d x = 3 4 . [RIGHT CLICK] [ASK AI] [ EXAMPLE ] Consider
∬ R x e x y d A , R = [ 2 , 4 ] × [ 1 , 3 ] . \iint_R xe^{xy}\,dA,
\qquad
R=[2,4]\times[1,3]. ∬ R x e x y d A , R = [ 2 , 4 ] × [ 1 , 3 ] . Integrating in y y y x x x
∬ R x e x y d A = ∫ 2 4 ∫ 1 3 x e x y d y d x = ∫ 2 4 [ e x y ] 1 3 d x = ∫ 2 4 ( e 3 x − e x ) d x = [ e 3 x 3 − e x ] 2 4 = e 12 3 − e 4 − e 6 3 + e 2 . \begin{aligned}
\iint_R xe^{xy}\,dA
&=\int_2^4\int_1^3xe^{xy}\,dy\,dx\\
&=\int_2^4\left[e^{xy}\right]_{1}^{3}dx\\
&=\int_2^4(e^{3x}-e^x)\,dx\\
&=\left[\frac{e^{3x}}3-e^x\right]_2^4\\
&=\frac{e^{12}}3-e^4-\frac{e^6}{3}+e^2.
\end{aligned} ∬ R x e x y d A = ∫ 2 4 ∫ 1 3 x e x y d y d x = ∫ 2 4 [ e x y ] 1 3 d x = ∫ 2 4 ( e 3 x − e x ) d x = [ 3 e 3 x − e x ] 2 4 = 3 e 12 − e 4 − 3 e 6 + e 2 . The reverse order leaves x e x y xe^{xy} x e x y x x x
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] For R = [ − 1 , 1 ] × [ 0 , π / 4 ] R=[-1,1]\times[0,\pi/4] R = [ − 1 , 1 ] × [ 0 , π /4 ]
∬ R x 2 sin y d A = ( ∫ − 1 1 x 2 d x ) ( ∫ 0 π / 4 sin y d y ) = 2 3 ( 1 − 1 2 ) . \begin{aligned}
\iint_R x^2\sin y\,dA
&=\left(\int_{-1}^{1}x^2\,dx\right)
\left(\int_0^{\pi/4}\sin y\,dy\right)\\
&=\frac23\left(1-\frac1{\sqrt2}\right).
\end{aligned} ∬ R x 2 sin y d A = ( ∫ − 1 1 x 2 d x ) ( ∫ 0 π /4 sin y d y ) = 3 2 ( 1 − 2 1 ) . [RIGHT CLICK] [ASK AI] [ EXAMPLE ] Define on [ 0 , 1 ] 2 [0,1]^2 [ 0 , 1 ] 2
f ( x , y ) = { 1 x 2 , 0 < y ≤ x ≤ 1 , − 1 y 2 , 0 < x < y ≤ 1 , 0 , x = 0 or y = 0. f(x,y)=
\begin{cases}
\dfrac1{x^2},&0<y\le x\le1,\\[6pt]
-\dfrac1{y^2},&0<x<y\le1,\\[6pt]
0,&x=0\text{ or }y=0.
\end{cases} f ( x , y ) = ⎩ ⎨ ⎧ x 2 1 , − y 2 1 , 0 , 0 < y ≤ x ≤ 1 , 0 < x < y ≤ 1 , x = 0 or y = 0. The values on the axes remove the undefined endpoint without changing any
integral. For fixed x > 0 x>0 x > 0 y y y y = x y=x y = x
∫ 0 1 f ( x , y ) d y = ∫ 0 x 1 x 2 d y + ∫ x 1 − 1 y 2 d y = 1 x + ( 1 − 1 x ) = 1. \begin{aligned}
\int_0^1 f(x,y)\,dy
&=\int_0^x\frac1{x^2}\,dy
\;+\int_x^1-\frac1{y^2}\,dy\\
&=\frac1x+\left(1-\frac1x\right)
=1.
\end{aligned} ∫ 0 1 f ( x , y ) d y = ∫ 0 x x 2 1 d y + ∫ x 1 − y 2 1 d y = x 1 + ( 1 − x 1 ) = 1. Thus one iterated integral gives
∫ 0 1 ∫ 0 1 f ( x , y ) d y d x = 1. \int_0^1\int_0^1f(x,y)\,dy\,dx=1. ∫ 0 1 ∫ 0 1 f ( x , y ) d y d x = 1. For fixed y > 0 y>0 y > 0 x x x x = y x=y x = y
∫ 0 1 f ( x , y ) d x = ∫ 0 y − 1 y 2 d x + ∫ y 1 1 x 2 d x = − 1 y + ( 1 y − 1 ) = − 1. \begin{aligned}
\int_0^1f(x,y)\,dx
&=\int_0^y-\frac1{y^2}\,dx
\;+\int_y^1\frac1{x^2}\,dx\\
&=-\frac1y+\left(\frac1y-1\right)
=-1.
\end{aligned} ∫ 0 1 f ( x , y ) d x = ∫ 0 y − y 2 1 d x + ∫ y 1 x 2 1 d x = − y 1 + ( y 1 − 1 ) = − 1. The other order gives
∫ 0 1 ∫ 0 1 f ( x , y ) d x d y = − 1. \int_0^1\int_0^1f(x,y)\,dx\,dy=-1. ∫ 0 1 ∫ 0 1 f ( x , y ) d x d y = − 1. There is no contradiction. The positive and negative parts each have
infinite integral near the origin, so f f f
[RIGHT CLICK] [ASK AI] Fubini and iterated integrals Builds on Double integrals and Riemann sums .