Please help with this messy integral problem,

[econan1214]

so I've come down to the part where I have the final integral to integrate on, which is:

double integrals: //(9u/19 + 10v/19)(15/19) dudv, [region of integration is : 10/3 < u < 29/3, -3<v<23/5 ]

This seems to be simple, but the fractions got me all messed up. I've gone through all the process and got the answer of 101, while wolframalpha gives 133.

By taking the antiderivatives of du, and substituting, I have :

intergral of (12615/722 + 1450v/361) - (750/361+500v/361)dv,
simplyfy: (585/38 - 50v/19)dv
taking the anti derivatives:

(585v/38 - 50v^2/38) from -3<v<23/5,

after substituting and simplyfying, i got 101.
but wolframalpha has 133.

Please let me know which one is correct. thanks

 

[stapel]

so I've come down to the part where I have the final integral to integrate on, which is:

double integrals: //(9u/19 + 10v/19)(15/19) dudv, [region of integration is : 10/3 < u < 29/3, -3<v<23/5 ]

Please reply with confirmation or correction:

. . . . .\(\displaystyle \displaystyle \int_{-3}^{\frac{23}{5}}\, \int_{\frac{10}{3}}^{\frac{29}{3}}\, \left(\dfrac{15}{19}\right)\, \left(\dfrac{9u}{19}\, +\, \dfrac{10v}{19}\right)\, du\, dv\)

I've gone through all the process and got the answer of 101, while wolframalpha gives 133.

By taking the antiderivatives of du, and substituting, I have :

intergral of (12615/722 + 1450v/361) - (750/361+500v/361)dv,
simplyfy: (585/38 - 50v/19)dv

How are you getting this? Treating \(\displaystyle \,v\,\) as a constant, I'm getting this:

. . . . .\(\displaystyle \displaystyle \int_{\frac{10}{3}}^{\frac{29}{3}}\, \left(\dfrac{15}{19}\right)\, \left(\dfrac{9u}{19}\, +\, \dfrac{10v}{19}\right)\, du\, =\, \int_{\frac{10}{3}}^{\frac{29}{3}}\, \left[\, \left(\dfrac{135}{361}\right)u\, +\, \left(\dfrac{150}{361}\right)v\, \right]\, du\)

. . . . .\(\displaystyle \displaystyle =\, \left[\, \left(\dfrac{135}{361}\right)\, \left(\dfrac{u^2}{2}\right)\, +\, \left(\dfrac{150}{361}\right)vu\, \right]\, \bigg|_{\frac{10}{3}}^{\frac{29}{3}}\)

. . . . .\(\displaystyle \displaystyle =\, \left(\dfrac{135}{722}\right)\, \left(\left(\dfrac{29}{3}\right)^2\, -\, \left(\dfrac{10}{3}\right)^2\right)\, +\, \left(\dfrac{150}{361}\, v\right)\, \left(\dfrac{29}{3}\, -\, \dfrac{10}{3}\right)\)

. . . . .\(\displaystyle \displaystyle =\, \left(\dfrac{15\, \cdot\, 9}{19\, \cdot\, 38}\right)\, \left(\dfrac{19\, \cdot\, 39}{9}\right)\, +\, \left(\dfrac{50\, \cdot\, 3}{19\, \cdot\, 19}\, v\right)\, \left(\dfrac{19}{3}\right)\)

. . . . .\(\displaystyle \displaystyle =\, \dfrac{585}{38}\, +\, \dfrac{50}{19}\, v\, =\, \dfrac{5}{19}\, \left(\dfrac{117}{2}\, +\,10v\right) \)

Wolfram Alpha concurs. (here) And doing the remaining integration then leads to the expected result. (here) Did the signs maybe get messed up somewhere in your working...?