Radius for 𝑓(𝑥)=∑𝑘2𝑥2 𝑅=lim𝑘sup𝑘√|𝑘2|⟹𝑅=1 because 𝑘2𝑘≥1⟹𝑅≥1 𝑅≤1 because (1+𝜀)𝑘≥(1+𝜀𝑘)⟹(∃𝑘0)(∀𝑘≥𝑘0)(1+𝜀)𝑘≥(1+𝑘𝜀)≥𝑘2 □ Riemann integral Calculate integral of lim𝑎⟶1∑𝑐𝑜𝑠(𝑎𝑗2)𝑎𝑗(1−𝑎) so mesh is like: 𝑥0=1,𝑥1=𝑎,𝑥𝑗=𝑎𝑗,𝑥𝑁=0 𝑅∞(𝑓)=∑∞𝑗=0𝑓(𝑎𝑗)(𝑥𝑗−𝑥𝑗+1) where 𝑥𝑗−𝑥𝑗+1=𝑎𝑗(1−𝑎) take 𝑥𝑗=𝑎𝑗=𝜉𝑗 so 𝑅∞(𝑓,𝜉)=∫10𝑓(𝑡)𝑑𝑡=∫10cos(√𝑥)𝑑𝑥 □