Упр.370 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

Изобразить фигуру, площадь которой равна данному интегралу, и вычислить эту площадь:

1. $$\int_{-2}^{-\frac{3}{2}}(-2-3x)\,dx$$;
2. $$\int_{-2}^{-3}(6-2x)\,dx$$;
3. $$\int_{0}^{4}(12+x-x^2)\,dx$$;
4. $$\int_{-2}^{0}(x^2-3x)\,dx$$;
5. $$\int_{2\pi}^{3\pi}\sin x\,dx$$;
6. $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos x\,dx$$;
7. $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{2}}\cos 2x\,dx$$;
8. $$\int_{\frac{\pi}{6}}^{\pi}\sin x\,dx$$.

1. $$\int_{-2}^{-\frac32}(-2-3x)\,dx=\left(-2x-\frac32x^2\right)\Bigg|_{-2}^{-\frac32}$$

   $$=\left(3-\frac{27}{8}\right)-\left(4-6\right)=5-\frac{27}{8}=1\frac58.$$

2. $$\int_{-2}^{3}(6-2x)\,dx=\left(6x-x^2\right)\Bigg|_{-2}^{3}$$

   $$=(18-9)-(-12-4)=9+16=25.$$

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

   $$=48+8-\frac{64}{3}=\frac{104}{3}=34\frac23.$$

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

   $$=0-\left(-\frac83-6\right)=\frac83+6=\frac{26}{3}=8\frac23.$$

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

   $$=-\cos 3\pi+\cos 2\pi=1+1=2.$$

6. $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos x\,dx=\sin x\Bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$$

   $$=\sin\frac{\pi}{2}-\sin\left(-\frac{\pi}{2}\right)=1-(-1)=2.$$

7. $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{2}}\cos x\,dx=\sin x\Bigg|_{-\frac{\pi}{4}}^{\frac{\pi}{2}}$$

   $$=\sin\frac{\pi}{2}-\sin\left(-\frac{\pi}{4}\right)=1+\frac{\sqrt2}{2}.$$

8. $$\int_{\frac{\pi}{6}}^{\pi}\sin x\,dx=-\cos x\Bigg|_{\frac{\pi}{6}}^{\pi}$$

   $$=-\cos\pi+\cos\frac{\pi}{6}=1+\frac{\sqrt3}{2}.$$

Ответ

1) $$1\frac58$$; 2) $$25$$; 3) $$\frac{104}{3}$$; 4) $$\frac{26}{3}$$; 5) $$2$$; 6) $$2$$; 7) $$1+\frac{\sqrt2}{2}$$; 8) $$1+\frac{\sqrt3}{2}$$.