The value of limn→∞{(n3+1)(n3+23)(n3+33)........(n3+n3)n3}1n is equal to α.eβ∫x31+x3dx, where αϵN,βϵR,then find α−β.
y=limn→∞{(n3+1)(n3+23)(n3+33)........(n3+n3)n3}1n
Taking log on both sides, we get
logy=limn→∞1n{∑nr=1log(n3+r3)−logn3}⇒logy=limn→∞1n∑nr=1log[1+(rn)3]⇒logy=∫10log(1+x3)dx⇒logy=[log(1+x3).x]10−∫103x21+x3xdx⇒logy=log2−3∫10x21+x3xdx ⇒y=2.e−3∫10x31+x3∴α−β=2−(−3)=5