To evaluate : ∫042sds
∫042sds=15ln2
Consider, ∫042sds=[2sln2]04
(Since, we have ∫axdx=axlna+C)
∫042sds=24ln2−20ln2 (Since 24=16;20=1)
=16ln2−1ln2=15ln2
Conclusion: ∫042sds=15ln2