Ex 7.9, 15 - Direct Integrate x ex2 dx from 0 to 1 - Definate Integration - By Substitution

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Ex7.9, 15 ∫_0^1β–’γ€–π‘₯ 〖𝑒^π‘₯γ€—^2 γ€— 𝑑π‘₯ Step 1 :- Let F(π‘₯)=∫1β–’γ€–π‘₯ 𝑒^(π‘₯^2 ) 𝑑π‘₯γ€— Let π‘₯^2=𝑑 Differentiating w.r.t.π‘₯ 𝑑/𝑑π‘₯ (π‘₯^2 )=𝑑𝑑/𝑑π‘₯ 2π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/2π‘₯ Therefore, ∫1β–’γ€–π‘₯ 𝑒^(π‘₯^2 ) 𝑑π‘₯=∫1β–’γ€–π‘₯ 𝑒^𝑑 𝑑𝑑/2π‘₯γ€—γ€— =1/2 ∫1▒〖𝑒^𝑑 𝑑𝑑〗 =1/2 𝑒^𝑑 Putting 𝑑=π‘₯^2 =1/2 𝑒^(π‘₯^2 ) Hence F(π‘₯)=1/2 𝑒^(π‘₯^2 ) Step 2 :- ∫_0^1β–’γ€–π‘₯𝑒^π‘₯ 𝑑π‘₯γ€—=𝐹(1)βˆ’πΉ(0) =1/2 𝑒^(1^2 )βˆ’1/2 𝑒^(0^2 ) =1/2 𝑒^1βˆ’1/2 𝑒^0 =1/2 (π‘’βˆ’1)

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