Một bài tích phân suy luận hay

Tìm \(f(9)\), biết rằng\(\int\limits_0^{{x^2}} {f\left( t \right)dt} = x\cos \left( {\pi x} \right)\)

A.\(f\left( 9 \right) = – \dfrac{1}{6}\)

B.\(f\left( 9 \right) = \dfrac{1}{6}\)

C. \(f\left( 9 \right) = – \dfrac{1}{9}\)

D. \(f\left( 9 \right) = \dfrac{1}{9}\)

Nếu \(F’\left( x \right) = f\left( x \right)\) và \(g\left( x \right) = \int\limits_a^x {f\left( t \right)dt}\)

thì \(g\left( x \right) = F\left( x \right) – F\left( a \right) \Rightarrow g’\left( x \right) = F’\left( x \right) \Rightarrow g’\left( x \right) = f\left( x \right)\) \(\begin{array}{l} g\left( {{x^2}} \right) = F\left( {{x^2}} \right) – F\left( a \right) \Rightarrow g’\left( {{x^2}} \right) = F’\left( {{x^2}} \right) \Rightarrow g’\left( {{x^2}} \right) = 2xf\left( {{x^2}} \right)\\ \Leftrightarrow 2xf\left( {{x^2}} \right) = {\left[ {x\cos \left( {\pi x} \right)} \right]^\prime } = \cos \left( {\pi x} \right) – \pi x\sin \left( {\pi x} \right)\\ \Rightarrow 6f\left( 9 \right) = 1\\ \Rightarrow f\left( 9 \right) = \dfrac{1}{6} \end{array}\)