Question

calculate double integral xy^2/x^2+1 where R ={(x,y)/0<=x<=1,-3<=y<=3}

Answer 1

\[\int\limits_{}^{}\int\limits \frac{x y^2}{x^2+1} \, dxdy\]??????

Answer 2

limits of x=[0,1] and y=[-3,3]

Answer 3

do the inner first
\[\int\limits \frac{x y^2}{x^2+1} \, dx\]

Answer 4

\[\begin{array}{l} \text{Factor out constants:} \\ \text{}=y^2\int\limits \frac{x}{x^2+1} \, dx \\\end{array}\]

Answer 5

u-sub

u=x^2+1
du=2xdx

Answer 6

\[\frac{y^2}{2}\int\limits \frac{1}{u} \, du\]
\[ [ \frac{1}{2} y^2 \ln (u)]_{}^{}\]
\[\frac{1}{2} y^2 \ln \left(x^2+1\right)|_{0}^{1}\]

Answer 7

\[\int\limits_{-3}^{3}\frac{1}{2} y^2 \ln \left(2\right)dy\]

Answer 8

\[\int\limits \frac{1}{2} y^2 \log (2) \, dy\]
\[\frac{1}{6} y^3 \log (2)|_{-3}^{3}\]

Answer 9

Solve

Answer 10

i am getting \[\int\limits_{-3}^{3} y^2/2 (\ln(2) -\ln (1))dy\]

Answer 11

There's some mistake in your calculations?
or maybe its inverted

\[\int\limits\limits_{}^{}\int\limits\limits \frac{x y^2}{x^2+1} \, dxdy \leftarrow \rightarrow \int\limits\limits_{}^{}\int\limits\limits \frac{x y^2}{x^2+1} \, dydx\]

Answer 12

final anwer is 9 ln2/ln1 ?

Answer 13

i meant 9ln2

Answer 14

yes

Answer 15

thanx