Question

Вычислить определенный интеграл

Answer 1

$30)\ \ \int\limits^1_0\, \dfrac{4x-1}{\sqrt{2x^2-x+4}}\, dx=\int\limits^1_0\, \dfrac{4x-1}{\sqrt{2\cdot (\, (x-\frac{1}{4})^2+\frac{31}{16}\, )}}\, dx=\Big[\ t=x-\dfrac{1}{4}\ \Big]=\\\\\\=\dfrac{1}{\sqrt2}\, \int\limits^{3/4}_{-1/4}\, \dfrac{4t\, dt}{\sqrt{t^2+\frac{31}{16}}}=\dfrac{2}{\sqrt2}\cdot \int\limits^{3/4}_{-1/4}\, \dfrac{2t\, dt}{\sqrt{t^2+\frac{31}{16}}}=\sqrt2\cdot 2\sqrt{t^2+\dfrac{31}{16}}\ \Big|_{-1/4}^{3/4}=$

$=2\sqrt2\cdot \Big(\sqrt{\dfrac{40}{16}}-\sqrt{\dfrac{32}{16}}\Big)=2\sqrt2\cdot \Big(\sqrt{\dfrac{5}{2}}-\sqrt2\Big)=2\cdot (\sqrt5-2)\ ;$

$31)\ \ \int\limits^1_0\, (2x+1)\cdot cos(x^2+x)\, dx=\Big[\ t=x^2+x\ ,\ dt=(2x+1)\, dx\ ,\ t_1=0,t_2=2\ \Big]=\\\\\\=\int\limits^2_0\, cost\, dt=sint\ \Big|_0^2=sin2-sin\, 0=sin2-0=sin2\ ;$

$32)\ \ \int\limits^{\pi /2}_0\, sinx\cdot cos^7x\, dx=-\int\limits^{\pi /2}_0\, cos^7x\cdot (-sinx\, dx)=-\int\limits^{\pi /2}_0\, cos^7x\cdot d(cosx)=\\\\\\=-\dfrac{cos^8x}{8}\, \Big|_0^{\pi /2}=-\dfrac{1}{8}\cdot \Big(cos^8\dfrac{\pi}{2}-cos^80\Big)=-\dfrac{1}{8}\cdot (0-1)=\dfrac{1}{8}$

Answer 2

Пошаговое объяснение:см. во вложении