Biết rằng hàm số \(f\left( x \right)=a{{x}^{2}}+bx+c\) thỏa mãn \(\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=-\frac{7}{2}\), \(\int\limits_{0}^{2}{f\left( x \right)\text{d}x}=...

* Hướng dẫn giải

Ta có \(\int\limits_{0}^{d}{f\left( x \right)\text{d}x}=\left. \left( \frac{a}{3}{{x}^{3}}+\frac{b}{2}{{x}^{2}}+cx \right) \right|_{0}^{d}=\frac{a}{3}{{d}^{3}}+\frac{b}{2}{{d}^{2}}+cd\).

Do đó: \(\left\{ \begin{array}{l} \int\limits_0^1 {f\left( x \right){\rm{d}}x} = - \frac{7}{2} \Leftrightarrow \frac{a}{3} + \frac{b}{2} + c = - \frac{7}{2}\\ \int\limits_0^2 {f\left( x \right){\rm{d}}x} = - 2 \Leftrightarrow \frac{8}{3}a + 2b + 2c = - 2\\ \int\limits_0^3 {f\left( x \right){\rm{d}}x} = \frac{{13}}{2} \Leftrightarrow 9a + \frac{9}{2}b + 3c = \frac{{13}}{2} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = 1\\ b = 3\\ c = - \frac{{16}}{3} \end{array} \right.\).

Vậy \(P=a+b+c=-\frac{4}{3}\)