Question

решите..............

Answer 1

$intlimits^1_{-1} {(0.25x+3)^3} , dx = intlimits^1_{-1} { frac{1}{64}x^3+ frac{9}{16}x^2+ frac{27}{4}x+27 } , dx = \ =(frac{x^4}{4cdot 64}+ frac{9x^3}{3cdot16}+ frac{27x^2}{2cdot4}+27x)|^1_{-1}= \ =(frac{1^4}{256}+ frac{9cdot1^3}{48}+ frac{27cdot1^2}{8}+27)-(frac{(-1)^4}{256}+ frac{9cdot(-1)^3}{48}+ frac{27cdot(-1)^2}{8}+27cdot(-1))= \ =frac{1}{256}+ frac{9}{48}+ frac{27}{8}+27-frac{1}{256}+ frac{9}{48}-frac{27}{8}+27= frac{18}{48}+54=54 frac{3}{8}$