Question

помогите решить интегралы 6 штук​

Answer 1

$1)\ \ \int\limits^1_{-3}\dfrac{dx}{x^6}=\dfrac{x^{-5}}{-5}\Big|_{-3}^1=-\dfrac{1}{5}\cdot (1-\dfrac{1}{243})=-\dfrac{242}{1215}$

$2)\ \ \int\limits^1_{-2}\Big((x-1)^3+(x+2)^2\Big)\, dx=\Big(\dfrac{(x-1)^4}{4}+\dfrac{(x+2)^3}{3}\Big)\Big|_{-2}^1=\\\\\\=\dfrac{1}{4}\cdot (0+81)+\dfrac{1}{3}\cdot (27-0)=20,25+9=29,25$

$3)\ \ \int\limits^{\pi /2}_{-\pi /2} \, cos2x\, dx=\dfrac{1}{2}\cdot sin2x\Big|_{-\pi /2}^{\pi /2}=\dfrac{1}{2}\cdot (sin\pi - sin(-\pi ))=0$

$4)\ \ \int\limits^7_2\, \Big(\sqrt{x+2}-\dfrac{1}{x+3}\Big)\, dx=\Big(\dfrac{1}{2\sqrt{x+2}}-ln|x+3|\Big)\Big|_2^7=\\\\\\=\dfrac{1}{2\cdot 3}-\dfrac{1}{2\cdot 2}-(ln10-ln5)=\dfrac{1}{6}-\dfrac{1}{4}-ln2=-\dfrac{1}{12}-ln2$

$5)\ \ \int\limits^{ln3}_0\, e^{3x}\, dx=\dfrac{1}{3}\cdot e^{3x}\ \Big|_0^{ln3}=\dfrac{1}{3}\cdot (3^{3ln3}-e^0)=\dfrac{1}{3}\cdot (27-1)=\dfrac{26}{3}$

$6)\ \ \int\limits^2_0\, (x-2)(x^2-4x+5)\, dx=\int\limits^2_0\, (x^3-4x^2+5x-2x^2+8x-10)\, dx=\\\\\\=\int\limits^2_0\, (x^3-6x^2+12x-10)\, dx=\Big(\dfrac{x^4}{4}-2x^3+6x^2-10x\Big)\Big|_0^2=4-16+24-20=-8$