Question

しろ

誰かこの問題の途中式を丁寧に解いてくれる方いませんか

(2log 2-e-2) dy

Answer 1

mo1

参考・概略です

　∫[{2^(y)}・{log(2)}－{e^(y-2)}]dy

●∫[f(x)±g(x)]dx＝∫[f(x)]dx±∫[g(x)]dx　より

＝∫[{2^(y)}・{log(2)}]dy－∫[e^(y-2)]dy

●係数{log(2)}を前に出して

＝{log(2)}・∫[2^(y)]dy－∫[e^(y-2)}]dy

●∫[2^(y)]dy＝{2^(y)/log(2)}＋Ｃ　…　①
　∫[e^(y-2)}]dy＝{e^(y-2)}＋Ｃ　　…　②

＝{log(2)}・{2^(y)/log(2)}－{e^(y-2)}＋Ｃ

●log(2)を約分

＝{2^(y)}－{e^(y-2)}＋Ｃ