Connectedness 3

Using a computer program, I found the number of graphs on seven vertices that have $e$ edges and $k$ connected components: $\begin{array}{c||c|c|c|c|c|c|c} e \backslash k & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \hline 1 & 0 & 0 & 0 & 0 & 0 & 21 & 0 \\ \hline 2 & 0 & 0 & 0 & 0 & 210 & 0 & 0 \\ \hline 3 & 0 & 0 & 0 & 1295 & 35 & 0 & 0 \\ \hline 4 & 0 & 0 & 5250 & 735 & 0 & 0 & 0 \\ \hline 5 & 0 & 13377 & 6762 & 210 & 0 & 0 & 0 \\ \hline 6 & 16807 & 32417 & 5005 & 35 & 0 & 0 & 0 \\ \hline 7 & 68295 & 45360 & 2625 & 0 & 0 & 0 & 0 \\ \hline 8 & 156555 & 45990 & 945 & 0 & 0 & 0 & 0 \\ \hline 9 & 258125 & 35595 & 210 & 0 & 0 & 0 & 0 \\ \hline 10 & 331506 & 21189 & 21 & 0 & 0 & 0 & 0 \\ \hline 11 & 343140 & 9576 & 0 & 0 & 0 & 0 & 0 \\ \hline 12 & 290745 & 3185 & 0 & 0 & 0 & 0 & 0 \\ \hline 13 & 202755 & 735 & 0 & 0 & 0 & 0 & 0 \\ \hline 14 & 116175 & 105 & 0 & 0 & 0 & 0 & 0 \\ \hline 15 & 54257 & 7 & 0 & 0 & 0 & 0 & 0 \\ \hline 16 & 20349 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 17 & 5985 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 18 & 1330 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 19 & 210 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 20 & 21 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 21 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline \sum & 1866256 & 207536 & 20818 & 2275 & 245 & 21 & 1 \end{array}$

Let $P_k(p)$ be the probability that there are $k$ connected components. Then $\begin{aligned} P_1(p) &= 16807p^6 (1 - p)^{15} + 68295p^7 (1 - p)^{14} + 156555p^8 (1 - p)^{13} + 258125p^9 (1 - p)^{12} \\ &\quad + 331506p^{10} (1 - p)^{11} + 343140p^{11} (1 - p)^{10} + 290745p^{12} (1 - p)^9 + 202755p^{13} (1 - p)^8 \\ &\quad + 116175p^{14} (1 - p)^7 + 54257p^{15} (1 - p)^6 + 20349p^{16} (1 - p)^5 + 5985p^{17} (1 - p)^4 \\ &\quad + 1330p^{18} (1 - p)^3 + 210p^{19} (1 - p)^2 + 21p^{20} (1 - p) + p^{21}, \\ P_2(p) &= 13377p^5 (1 - p)^{16} + 32417p^6 (1 - p)^{15} + 45360p^7 (1 - p)^{14} + 45990p^8 (1 - p)^{13} \\ &\quad + 35595p^9 (1 - p)^{12} + 21189p^{10} (1 - p)^{11} + 9576p^{11} (1 - p)^{10} + 3185p^{12} (1 - p)^9 \\ &\quad + 735p^{13} (1 - p)^8 + 105p^{14} (1 - p)^7 + 7p^{15} (1 - p)^6, \\ P_3(p) &= 5250p^4 (1 - p)^{17} + 6762p^5 (1 - p)^{16} + 5005p^6 (1 - p)^{15} + 2625p^7 (1 - p)^{14} \\ &\quad + 945p^8 (1 - p)^{13} + 210p^9 (1 - p)^{12} + 21p^{10} (1 - p)^{11}, \\ P_4(p) &= 1295p^3 (1 - p)^{18} + 735p^4 (1 - p)^{17} + 210p^5 (1 - p)^{16} + 35p^6 (1 - p)^{15}, \\ P_5(p) &= 210p^2 (1 - p)^{19} + 35p^3 (1 - p)^{18}, \\ P_6(p) &= 21p (1 - p)^{20}, \\ P_7(p) &= (1 - p)^{21}. \end{aligned}$

We can compute that $\begin{aligned} \int_0^1 P_1(p) \ dp &= \frac{242565}{369512}, \\ \int_0^1 P_2(p) \ dp &= \frac{211093}{2217072}, \\ \int_0^1 P_3(p) \ dp &= \frac{859}{14212}, \\ \int_0^1 P_4(p) \ dp &= \frac{505}{10032}, \\ \int_0^1 P_5(p) \ dp &= \frac{39}{836}, \\ \int_0^1 P_6(p) \ dp &= \frac{1}{22}, \\ \int_0^1 P_7(p) \ dp &= \frac{1}{22}. \end{aligned}$

Finally, $N(p) = P_1(p) + 2P_2(p) + 3P_3(p) + \dots + 7P_7(p),$ so $\begin{aligned} \int_0^1 N(p) \ dp &= \int_0^1 [P_1(p) + 2P_2(p) + 3P_3(p) + \dots + 7P_7(p)] \ dp \\ &= \int_0^1 P_1(p) \ dp + 2 \int_0^1 P_2(p) \ dp + 3 \int_0^1 P_3(p) \ dp + \dots + 7 \int_0^1 P_7(p) \ dp \\ &= \frac{242565}{369512} \cdot 1 + \frac{211093}{2217072} \cdot 2 + \frac{859}{14212} \cdot 3 \\ &\quad + \frac{505}{10032} \cdot 4 + \frac{39}{836} \cdot 5 + \frac{1}{22} \cdot 6 + \frac{1}{22} \cdot 7 \\ &= \frac{59911}{29172}. \end{aligned}$

@Christopher Criscitiello , is there any way to do these problems other than brute force?