Answer:
2
Step-by-step explanation:
[tex]\int\limits^4_0 {p/\sqrt{9+p^2} } \, dp[/tex]
We will use a u substitution
Let u = 9+p^2
du = 2p dp
The lower limit becomes
u = 9+0^2 = 9
The upper limit becomes
u = 9+4^2 = 25
[tex]\int\limits^4_0 {1/2*2p/\sqrt{9+p^2} } \, dp[/tex]
[tex]\int\limits^a_9 { 1/2\sqrt{u} } \, du[/tex] where a is 25
[tex]1/2 \int\limits^a_9 { 1/\sqrt{u} } \, du[/tex]
We know the intergral of 1 / sqrt(u) is sqrt(u)/ 1/2
1/2 * sqrt(u)/(1/2) evaluated at 25 and 9
sqrt(u) where u is25 - sqrt(u) where u is 9
sqrt(25) - sqrt(9)
5-3
2
