carilah nilai f(x) jika b. f'(x)= √x + 2/√x dan f(1) =1

Jawaban yang benar adalah f(x) = (2/3) x√(x) + 4√(x) - 11/3 Rumus integral : ∫ ax^(n) dx = (a/(n+1)) · x^(n+1) + C Sifat eksponen : a^(-n) = 1/(aⁿ) a^(m/n) = ⁿ√(a^(m)) a^(b+c) = a^(b) · a^(c) Pembahasan : f'(x)= √x + 2/√x f(x) = ∫ f'(x) dx = ∫ (√x + 2/√x) dx = ∫ (x^(1/2) + 2/(x^(1/2))) dx = ∫ (x^(1/2) + 2·x^(-1/2)) dx = 1/(1/2 +1) · x^(1/2 + 1) + (2/(-1/2 + 1)) · x^(-1/2 + 1) + c = 1/(1/2 + 2/2) · x^(1/2) · x^(1) + 2/(-1/2 + 2/2) · x^(-1/2 + 2/2) + c = 1/(3/2) x · x^(1/2)+ 2/(1/2) · x^(1/2) + c = 1 · 2/3 · x√(x) + 2 · 2/1 · √(x) + c = (2/3) x√(x) + 4√(x) + c f(1) = 1 (2/3) · 1√(1) + 4√(1) + c = 1 (2/3) + 4 + c = 1 c = 1 - 4 - (2/3) c = -3 - 2/3 c = -9/3 - 2/3 c = -11/3 Diperoleh : f(x) = (2/3) x√(x) + 4√(x) - 11/3 Jadi nilai fungsi f(x) = (2/3) x√(x) + 4√(x) - 11/3