# trouble with double/iterated integration of ∫10[∫10v(u+v2)4du]dv\int^1_0[\int^1_0v(u+v^2)^4du]dv

I have:

For ∫10v(u+v2)4du\int^1_0v(u+v^2)^4du:

u substitution (using x instead since there’s a u in there already) with x=(u+v2),dx/du=1x=(u+v^2), dx/du=1

v∫10×4=v15x5=v(u+v2)55−v(0+v2)55|10=(v+v3)55−v155v\int^1_0x^4=v\frac{1}{5}x^5=v\frac{(u+v^2)^5}{5}-\frac{v(0+v^2)^5}{5}|^1_0=\frac{(v+v^3)^5}{5}-\frac{v^{15}}{5}

then

15[∫10(v+v3)5dv−∫10v15dv] \frac{1}{5}[\int^1_0(v+v^3)^5dv-\int^1_0v^{15}dv]

used substitution again with x=(v+v3),dx/du=(1+3v2)x= (v+v^3),dx/du=(1+3v^2) for the first one

∫10x5dx(1+3v2)−>11+3v2∫10x5dx=11+3v216x6|10−>11+3v216(v+v3)6|10\int^1_0\frac{x^5dx}{(1+3v^2)}->\frac{1}{1+3v^2}\int^1_0x^5dx=\frac{1}{1+3v^2}\frac{1}{6}x^6|^1_0->\frac{1}{1+3v^2}\frac{1}{6}(v+v^3)^6|^1_0

=124∗1−0= \frac{1}{24}*1-0

for the second:

∫10v15=116v16|10=116\int^1_0 v^{15}=\frac{1}{16}v^{16}|^1_0=\frac{1}{16}

15[124−116]=−1240\frac{1}{5}[\frac{1}{24}-\frac{1}{16}] = -\frac{1}{240}

The answer is supposed to be 31/30.

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Note that v(1+v2)5≠(v+v3)5v(1+v^2)^5\ne (v+v^3)^5. Rather, we have upon integrating with respect to uu,

∫10v(u+v2)4du=v5((1+v2)5−v10) \int_0^1 v(u+v^2)^4\,du=\frac v5 \left( (1+v^2)^5-v^{10} \right)

Now, note that the integration over vv is facilitated by the substitution w=1+v2w=1+v^2. This is left as an exercise for the reader.

An alternate approach would be to change the order of integration. This will greatly reduce the difficulty of the integral. As the limits of the inner integral do not depend upon vv this trivially becomes:

∫10[∫10v(u+v2)4 dv] du\int^1_0[\int^1_0v(u+v^2)^4\ dv]\ du

=∫10[110((u+v2)5)10] du=\int^1_0\left[\frac{1}{10}\bigg((u+v^2)^5\bigg)_0^1\right]\ du

=∫10110[(u+1)5−(u+0)5] du=\int^1_0\frac{1}{10}\left[(u+1)^5-(u+0)^5\right]\ du

=110∫10(u+1)5−u5 du=\frac{1}{10}\int^1_0(u+1)^5-u^{5}\ du

=110(16(u+1)6−16u6)10=\frac{1}{10}\bigg(\frac{1}{6}(u+1)^6-\frac{1}{6}u^6\bigg)_0^1

=\frac{1}{10}\bigg(\frac{1}{6}(1+1)^6-\frac{1}{6}1^6-\left(\frac{1}{6}(1+0)^6-\frac{1}{6}0^6\right)\bigg)_0^1=\frac{1}{10}\bigg(\frac{1}{6}(1+1)^6-\frac{1}{6}1^6-\left(\frac{1}{6}(1+0)^6-\frac{1}{6}0^6\right)\bigg)_0^1

=\frac{31}{30}=\frac{31}{30}