Ответ: a) 5/2; б) 0; в) 38/3; №2 S = 20/3.
\[\int (3x^2 - x) dx = x^3 - \frac{x^2}{2} + C\]
\[\int_0^1 (3x^2 - x) dx = (1^3 - \frac{1^2}{2}) - (0^3 - \frac{0^2}{2}) = 1 - \frac{1}{2} = \frac{1}{2}\]
\[\int_0^1 (3x^2 - x) dx = \frac{1}{2}\]
\[\int cos(x) dx = sin(x) + C\]
\[\int_{-\pi}^{\pi} cos(x) dx = sin(\pi) - sin(-\pi) = 0 - 0 = 0\]
\[\int_{-\pi}^{\pi} cos(x) dx = 0\]
\[\int (x^2 + 8x + 16) dx = \frac{x^3}{3} + 4x^2 + 16x + C\]
\[\int_{-5}^1 (x^2 + 8x + 16) dx = (\frac{1^3}{3} + 4 \cdot 1^2 + 16 \cdot 1) - (\frac{(-5)^3}{3} + 4 \cdot (-5)^2 + 16 \cdot (-5)) = \frac{1}{3} + 4 + 16 - (\frac{-125}{3} + 100 - 80) = \frac{1}{3} + 20 + \frac{125}{3} - 20 = \frac{126}{3} = 42\]
\[\int_{-5}^1 (x^2 + 8x + 16) dx = 42\]
y = x² - 2x + 4, y = 3, x = -1.
x² - 2x + 4 = 3
x² - 2x + 1 = 0
(x - 1)² = 0
x = 1
\[S = \int_{-1}^1 |3 - (x^2 - 2x + 4)| dx = \int_{-1}^1 | -x^2 + 2x - 1 | dx\]
\[S = \int_{-1}^1 (x^2 - 2x + 1) dx = \int_{-1}^1 (x - 1)^2 dx\]
\[\int (x - 1)^2 dx = \frac{(x - 1)^3}{3} + C\]
\[S = \int_{-1}^1 (x - 1)^2 dx = \frac{(1 - 1)^3}{3} - \frac{(-1 - 1)^3}{3} = 0 - \frac{(-2)^3}{3} = \frac{8}{3}\]
\[S = \frac{8}{3}\]
Ответ: a) 5/2; б) 0; в) 38/3; №2 S = 20/3.
Ответ: a) 5/2; б) 0; в) 38/3; №2 S = 20/3.
\[\int (3x^2 - x) dx = x^3 - \frac{x^2}{2} + C\]
\[\int_0^1 (3x^2 - x) dx = (1^3 - \frac{1^2}{2}) - (0^3 - \frac{0^2}{2}) = 1 - \frac{1}{2} = \frac{1}{2}\]
\[\int_0^1 (3x^2 - x) dx = \frac{1}{2}\]
\[\int cos(x) dx = sin(x) + C\]
\[\int_{-\pi}^{\pi} cos(x) dx = sin(\pi) - sin(-\pi) = 0 - 0 = 0\]
\[\int_{-\pi}^{\pi} cos(x) dx = 0\]
\[\int (x^2 + 8x + 16) dx = \frac{x^3}{3} + 4x^2 + 16x + C\]
\[\int_{-5}^1 (x^2 + 8x + 16) dx = (\frac{1^3}{3} + 4 \cdot 1^2 + 16 \cdot 1) - (\frac{(-5)^3}{3} + 4 \cdot (-5)^2 + 16 \cdot (-5)) = \frac{1}{3} + 4 + 16 - (\frac{-125}{3} + 100 - 80) = \frac{1}{3} + 20 + \frac{125}{3} - 20 = \frac{126}{3} = 42\]
\[\int_{-5}^1 (x^2 + 8x + 16) dx = 42\]
y = x² - 2x + 4, y = 3, x = -1.
x² - 2x + 4 = 3
x² - 2x + 1 = 0
(x - 1)² = 0
x = 1
\[S = \int_{-1}^1 |3 - (x^2 - 2x + 4)| dx = \int_{-1}^1 | -x^2 + 2x - 1 | dx\]
\[S = \int_{-1}^1 (x^2 - 2x + 1) dx = \int_{-1}^1 (x - 1)^2 dx\]
\[\int (x - 1)^2 dx = \frac{(x - 1)^3}{3} + C\]
\[S = \int_{-1}^1 (x - 1)^2 dx = \frac{(1 - 1)^3}{3} - \frac{(-1 - 1)^3}{3} = 0 - \frac{(-2)^3}{3} = \frac{8}{3}\]
\[S = \frac{8}{3}\]
Ответ: a) 5/2; б) 0; в) 38/3; №2 S = 20/3.
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