Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 64

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\[\textbf{а)}\ y = x;\]

\[F(x) = \frac{x^{2}}{2};\]

\[\int_{0}^{1}\text{xdx} + \int_{1}^{3}\text{xdx} = \int_{0}^{3}\text{xdx};\]

\[\int_{0}^{3}\text{xdx} = F(3) - F(0) =\]

\[= \frac{3^{2}}{2} - \frac{0}{2} = \frac{9}{2} = 4,5.\]

\[\textbf{б)}\ y = \sin x;\]

\[F(x) = - \cos x;\]

\[\int_{0}^{\pi}{\sin x\text{dx}} + \int_{\pi}^{2\pi}{\sin x\text{dx}} =\]

\[= \int_{0}^{2\pi}{\sin x\text{dx}};\]

\[\int_{0}^{2\pi}{\sin x\text{dx}} = F(2\pi) - F(0) =\]

\[= - \cos{2\pi} - \left( - \cos 0 \right) = 0.\]

\[\textbf{в)}\ y = \frac{1}{x};\]

\[F(x) = \ln|x|;\]

\[\int_{1}^{2}\frac{\text{dx}}{x} + \int_{2}^{e}\frac{\text{dx}}{x} = \int_{1}^{e}\frac{\text{dx}}{x};\]

\[\int_{1}^{e}\frac{\text{dx}}{x} = F(e) - F(1) =\]

\[= \ln|e| - \ln|1| = 1 - 0 = 1.\]

\[\textbf{г)}\ y = e^{x};\]

\[F(x) = e^{x};\]

\[\int_{0}^{1}{e^{x}\text{dx}} + \int_{1}^{2}{e^{x}\text{dx}} + \int_{2}^{3}{e^{x}\text{dx}} =\]

\[= \int_{0}^{3}{e^{x}\text{dx}};\]

\[\int_{0}^{3}{e^{x}\text{dx}} = F(3) - F(0) =\]

\[= e^{3} - e^{0} = e^{3} - 1.\]

\[\textbf{д)}\ y = \sin x;\]

\[F(x) = - \cos x;\]

\[= \int_{0}^{\pi}{\sin x\text{dx}};\]

\[\int_{0}^{\pi}{\sin x\text{dx}} = F(\pi) - F(0) =\]

\[= - \cos\pi - \left( - \cos 0 \right) =\]

\[= - ( - 1) - ( - 1) = 2.\]