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Question Number 139587 by mathlove last updated on 29/Apr/21

Answered by qaz last updated on 29/Apr/21

$$f\left(2\right)={\int }_1^2{x}^2+3xdx+f\left(1\right)=(\frac{1}{3}{x}^3+\frac{3}{2}{x}^2){\mid }_1^2+5=\frac{71}{6}$$$$f\left(0\right)=f\left(1\right)-{\int }_0^{1}6x+4dx=9-(3x^2+4x){\mid }_0^1=2$$

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