Question

решите пжжжжжжжжжжжж​

Answer 1

$\int\limits^4_1 {2x^2} \, dx=\frac{2x^{2+1}}{2+1}|^4_1=\frac{2x^3}{3}|_1^4=\frac{2}{3}(4^3-1^3)=\frac{2}{3}(64-1) =\frac{2}{3}*63 =42\\\\\\\int\limits^{\pi/4}_0 {cosx} \, dx=sinx|^{\pi/4}_0=sin(\pi/4)-sin0=\frac{\sqrt{2}}{2}-0=\frac{\sqrt{2}}{2}\\\\\\\int\limits^0_{-1} {(3x^2-4x+2)} \, dx=(\frac{3x^3}{3}-\frac{4x^2}{2}+2x)|_{-1}^0=(x^3-2x^2+2x)|_{-1}^0=\\\\=(0-0+0)-((-1)^3-2(-1)^2+2(-1))=-(-1-2-2)=-(-5)=5$