=∫12loge1dx+∫23loge2dx++∫34loge3dx=0+(loge2)[x]23+(loge3)[x]34=∫12loge1dx+∫23loge2dx++∫34loge3dx=∫12loge[x]dx+∫23loge[x]dx++∫34loge[x]dx Thus, =(loge2)1+(loge3)1 Equivalent of =loge6 Thus, equivalent of y=m1x Equivalent of y=m2x Equivalent mass of 427418 Thus, percentage of 2.0×10−5/oC