Упр.1011 ГДЗ Алимов 10-11 класс (Алгебра)
- $$\int_{-\pi}^{\pi}\sin 2x\,dx$$
- $$\int_{0}^{\frac{\pi}{2}}\sin x\cos x\,dx$$
- $$\int_{0}^{\frac{\pi}{4}}(\cos 2x-\sin 2x)\,dx$$
- $$\int_{0}^{\pi}(\sin 4x+\cos 4x)\,dx$$
- $$\int_{0}^{3}x^2\sqrt{x+1}\,dx$$
- $$\int_{3}^{4}\frac{x^2-4x+5}{x-2}\,dx$$
1) $$\int\limits_{-\pi}^{\pi}\sin^2 x\,dx=\int\limits_{-\pi}^{\pi}\frac{1-\cos 2x}{2}\,dx=\left(\frac{x}{2}-\frac{\sin 2x}{4}\right)\Bigg|_{-\pi}^{\pi}=\pi.$$
2) $$\int\limits_{0}^{\pi/2}\sin x\cos x\,dx=\int\limits_{0}^{\pi/2}\frac{1}{2}\sin 2x\,dx=\left(-\frac{\cos 2x}{4}\right)\Bigg|_{0}^{\pi/2}=\frac{1}{2}.$$
3) $$\int\limits_{0}^{\pi/4}(\cos^2 x-\sin^2 x)\,dx=\int\limits_{0}^{\pi/4}\cos 2x\,dx=\left(\frac{\sin 2x}{2}\right)\Bigg|_{0}^{\pi/4}=\frac{1}{2}.$$
4) $$\int\limits_{0}^{\pi}(\sin^4 x+\cos^4 x)\,dx=\int\limits_{0}^{\pi}\left(1-2\sin^2 x\cos^2 x\right)\,dx$$
$$=\int\limits_{0}^{\pi}\left(1-\frac{1}{2}\sin^2 2x\right)\,dx=\int\limits_{0}^{\pi}\left(\frac{3}{4}+\frac{\cos 4x}{4}\right)\,dx=\left(\frac{3x}{4}+\frac{\sin 4x}{16}\right)\Bigg|_{0}^{\pi}=\frac{3\pi}{4}.$$
5) $$\int\limits_{0}^{3}x^2\sqrt{x+1}\,dx=\int\limits_{0}^{3}(x+1-1)^2\sqrt{x+1}\,dx$$
$$=\int\limits_{0}^{3}\left((x+1)^{5/2}-2(x+1)^{3/2}+(x+1)^{1/2}\right)\,dx$$
$$=\left(\frac{2}{7}(x+1)^{7/2}-\frac{4}{5}(x+1)^{5/2}+\frac{2}{3}(x+1)^{3/2}\right)\Bigg|_{0}^{3}=\frac{116}{15}.$$
6) $$\int\limits_{3}^{4}\frac{x^2-4x+5}{x-2}\,dx=\int\limits_{3}^{4}\left(x-2+\frac{1}{x-2}\right)\,dx$$
$$=\left(\frac{x^2}{2}-2x+\ln(x-2)\right)\Bigg|_{3}^{4}=\frac{3}{2}+\ln 2.$$
Ответ
1) $$\pi$$; 2) $$\frac{1}{2}$$; 3) $$\frac{1}{2}$$; 4) $$\frac{3\pi}{4}$$; 5) $$\frac{116}{15}$$; 6) $$\frac{3}{2}+\ln 2$$.
