Упр.23.13 ГДЗ Мордкович Семенов 11 класс (Алгебра)

а) (?/4,3?/4)?cos(x/3-?/12)dx; г) (?/12,?/4)?3sin(3x+?/4)dx;
б) (0,?/4)?2/sin^2(2x+?/4)dx; д) (-?/6,0)?1/(2cos^2(x/2-?/4))dx;
в) -(-?/3,?/3)?sin(x)cos(x)dx; е) (?/12,?/4)?(sin^2(x)-cos^2(x))dx. Найдите площадь криволинейной трапеции, ограниченной снизу осью абсцисс и сверху графиком функции y=f(x):
а) f(x)={x^3+1, -1×1; x^2-4x+5, 1

1. а)

   $$\int_{\pi/4}^{3\pi/4}\cos\left(\frac{x}{3}-\frac{\pi}{12}\right)\,dx =3\sin\left(\frac{x}{3}-\frac{\pi}{12}\right)\Bigg|_{\pi/4}^{3\pi/4}$$
   $$=3\sin\frac{\pi}{6}-3\sin 0=\frac{3}{2}.$$

2. б)

   $$\int_{0}^{\pi/4}\frac{2}{\sin^2\left(2x+\frac{\pi}{4}\right)}\,dx =-\ctg\left(2x+\frac{\pi}{4}\right)\Bigg|_{0}^{\pi/4}$$
   $$=-\ctg\frac{3\pi}{4}+\ctg\frac{\pi}{4}=1+1=2.$$

3. в)

   $$-\int_{-\pi/3}^{\pi/3}\sin x\cos x\,dx =-\frac12\int_{-\pi/3}^{\pi/3}\sin 2x\,dx =\frac14\cos 2x\Bigg|_{-\pi/3}^{\pi/3}=0.$$

4. г)

   $$\int_{\pi/12}^{\pi/4}3\sin\left(3x+\frac{\pi}{4}\right)\,dx =-\cos\left(3x+\frac{\pi}{4}\right)\Bigg|_{\pi/12}^{\pi/4}$$
   $$=-\cos\pi+\cos\frac{\pi}{2}=1.$$

5. д)

   $$\int_{-\pi/6}^{0}\frac{1}{2\cos^2\left(\frac{x}{2}-\frac{\pi}{4}\right)}\,dx =\tg\left(\frac{x}{2}-\frac{\pi}{4}\right)\Bigg|_{-\pi/6}^{0}$$
   $$=\tg\left(-\frac{\pi}{4}\right)-\tg\left(-\frac{\pi}{3}\right)=\sqrt{3}-1.$$

6. е)

   $$\int_{\pi/12}^{\pi/4}\left(\sin^2 x-\cos^2 x\right)\,dx =-\int_{\pi/12}^{\pi/4}\cos 2x\,dx =-\frac12\sin 2x\Bigg|_{\pi/12}^{\pi/4}$$
   $$=-\frac12\sin\frac{\pi}{2}+\frac12\sin\frac{\pi}{6} =-\frac12+\frac14=-\frac14.$$

Ход решения

1. а)

   $$S=\int_{-1}^{1}(x^3+1)\,dx+\int_{1}^{4}(x^2-4x+5)\,dx$$
   $$=\left(\frac{x^4}{4}+x\right)\Bigg|_{-1}^{1} +\left(\frac{x^3}{3}-2x^2+5x\right)\Bigg|_{1}^{4}$$
   $$=\left(\frac14+1-\frac14+1\right)+\left(\frac{64}{3}-32+20-\frac13+2-5\right)=8.$$

2. б)

   $$S=\int_{0}^{1}2x\,dx+\int_{1}^{2}\frac{2}{x^2}\,dx$$
   $$=x^2\Bigg|_{0}^{1}-\frac{2}{x}\Bigg|_{1}^{2}=1+( -1+2)=2.$$

3. в)

   $$S=\int_{-1}^{1}0{,}5^x\,dx+\int_{1}^{2}\frac{1}{2x}\,dx$$
   $$=\frac{0{,}5^x}{\ln 0{,}5}\Bigg|_{-1}^{1}+\frac12\ln|x|\Bigg|_{1}^{2}$$
   $$=\frac12\ln 2+\frac{3}{2\ln 2}.$$

4. г)

   $$S=\int_{1}^{2}x^3\,dx+\int_{2}^{4}(x^2-8x+20)\,dx$$
   $$=\frac{x^4}{4}\Bigg|_{1}^{2}+\left(\frac{x^3}{3}-4x^2+20x\right)\Bigg|_{2}^{4}$$
   $$=\frac{15}{4}+\frac{269}{12}-\frac{268}{12}=\frac{59}{12}.$$

5. д)

   $$S=\int_{-3}^{0}\sqrt{x+4}\,dx+\int_{0}^{1}2(x-1)^2\,dx$$
   $$=\frac{2}{3}(x+4)^{3/2}\Bigg|_{-3}^{0}+\frac{2}{3}(x-1)^3\Bigg|_{0}^{1}$$
   $$=\frac{2}{3}(8-1)+\frac{2}{3}(0-(-1))=\frac{14}{3}+\frac{2}{3}=\frac{16}{3}.$$

6. е)

   $$S=\int_{0}^{1}2^x\,dx+\int_{1}^{e}\frac{2}{x}\,dx$$
   $$=\frac{2^x}{\ln 2}\Bigg|_{0}^{1}+2\ln x\Bigg|_{1}^{e}$$
   $$=\frac{2-1}{\ln 2}+2=\;2+\frac{1}{\ln 2}.$$

Ответ

а) $$\frac{3}{2}$$; б) $$2$$; в) $$0$$; г) $$1$$; д) $$\sqrt{3}-1$$; е) $$-\frac14$$.
Площади: а) $$8$$; б) $$2$$; в) $$\frac12\ln 2+\frac{3}{2\ln 2}$$; г) $$\frac{59}{12}$$; д) $$\frac{16}{3}$$; е) $$2+\frac{1}{\ln 2}$$.