Indefinite IntegrationHard
Question
The value of ∫[f(x)g″(x) - f″(x)g(x)]dx is equal to
Options
A.
B.f′(x) g(x) - f(x) g′(x)
C.f(x) g′(x) - f′(x) g(x)
D.f(x) g′(x) + f′(x) g′(x)
Solution
∫(f(x)g″ - f″(x)g(x)) dx
= ∫f(x)g″(x) dx - ∫f″(x)g(x) dx
= f(x) g′(x) - ∫f′(x)g′(x) - ∫f″(x)g(x)
= f(x) g′(x) - f′(x) g(x) + ∫f″(x)g(x) - ∫f″(x)g(x)
= f(x) g′(x) - f′(x) g(x) + C
= ∫f(x)g″(x) dx - ∫f″(x)g(x) dx
= f(x) g′(x) - ∫f′(x)g′(x) - ∫f″(x)g(x)
= f(x) g′(x) - f′(x) g(x) + ∫f″(x)g(x) - ∫f″(x)g(x)
= f(x) g′(x) - f′(x) g(x) + C
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