Упр.392 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

1. 392. Вычислить интегралы:
   1) $$\int_{-1}^{2} 2\,dx$$;
   2) $$\int_{-2}^{2} (3-x)\,dx$$;
   3) $$\int_{1}^{3} (x^2-2x)\,dx$$;
   4) $$\int_{-1}^{1} (2x-3x^2)\,dx$$;
   5) $$\int_{1}^{8} \sqrt[3]{x}\,dx$$;
   6) $$\int_{1}^{2} \frac{dx}{x^3}$$;
   7) $$\int_{0}^{\frac{\pi}{2}} \sin x\,dx$$;
   8) $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x\,dx$$.

1. $$\int_{-1}^{2} 2\,dx = 2x\Big|_{-1}^{2} = 2\cdot 2 — 2\cdot(-1) = 4+2=6.$$

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

   $$=\left(3\cdot 2-\frac{2^2}{2}\right)-\left(3\cdot(-2)-\frac{(-2)^2}{2}\right)=\left(6-2\right)-\left(-6-2\right)=12.$$

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

   $$=\left(\frac{3^3}{3}-3^2\right)-\left(\frac{1^3}{3}-1^2\right)=(9-9)-\left(\frac13-1\right)=\frac23.$$

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

   $$=(1^2-1^3)-\left(({-1})^2-({-1})^3\right)=(1-1)-(1+1)=-2.$$

5. $$\int_{1}^{8} \sqrt[3]{x}\,dx = \int_{1}^{8} x^{1/3}\,dx = \frac{3}{4}x^{4/3}\Big|_{1}^{8}$$

   $$=\frac{3}{4}\cdot 8^{4/3}-\frac{3}{4}\cdot 1^{4/3}=\frac{3}{4}\cdot 16-\frac{3}{4}=12-\frac{3}{4}=11\frac14.$$

6. $$\int_{1}^{2} \frac{dx}{x^3}=\int_{1}^{2} x^{-3}\,dx=\left(-\frac{1}{2x^2}\right)\Big|_{1}^{2}$$

   $$=-\frac{1}{2\cdot 2^2}+\frac{1}{2\cdot 1^2}=-\frac18+\frac12=\frac38.$$

7. $$\int_{0}^{\pi/2} \sin x\,dx = -\cos x\Big|_{0}^{\pi/2} = -\cos\frac{\pi}{2}+\cos 0=1.$$

8. $$\int_{-\pi/2}^{\pi/2} \cos x\,dx = \sin x\Big|_{-\pi/2}^{\pi/2} = \sin\frac{\pi}{2}-\sin\left(-\frac{\pi}{2}\right)=2.$$

Ответ

1) $$6$$; 2) $$12$$; 3) $$\frac23$$; 4) $$-2$$; 5) $$11\frac14$$; 6) $$\frac38$$; 7) $$1$$; 8) $$2$$.