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Question-139587




Question Number 139587 by mathlove last updated on 29/Apr/21
Answered by qaz last updated on 29/Apr/21
f(2)=∫_1 ^2 x^2 +3xdx+f(1)=((1/3)x^3 +(3/2)x^2 )∣_1 ^2 +5=((71)/6)  f(0)=f(1)−∫_0 ^1 6x+4dx=9−(3x^2 +4x)∣_0 ^1 =2
$${f}\left(\mathrm{2}\right)=\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{3}{xdx}+{f}\left(\mathrm{1}\right)=\left(\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \right)\mid_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{5}=\frac{\mathrm{71}}{\mathrm{6}} \\ $$$${f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{6}{x}+\mathrm{4}{dx}=\mathrm{9}−\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{2} \\ $$

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