20250701

$$ \int x^3 dx = \frac{1}{4}x^4 + C\ \int (10x-5) dx = 5x^2 - 5x + C\ \int 3x^2 dx = x^3 + C\ \int (3t^2 + 6t) dt = t^3 + 3t^2 + C $$

以下の関数を積分せよ：

$$ f(x) = 3x^2\ f(x) = 10x^4 + 8x\ f(x) = 12x^2 +15x^2 -1\ f(x) = - \frac{2}{x^2}\ f(x) = - \frac{6}{x^3}\ f(x) = 2 + \frac{4}{x^2}\ f(x) = \frac{1}{2 \sqrt{x}}\ f(x) = \frac{3}{2 \sqrt{x}} - 1\ f(x) = \frac{1}{2 \sqrt{x}} - \frac{3}{2 \sqrt{x}} $$

定積分

$$ \begin{align} \int^3_0 (4x - 3)dx &= \left[2x^2 -3x\right]^3_0 \ &= (2 \cdot 3^2 - 3 \cdot 3) - (2 \cdot 0^2 - 3 \cdot 0) \ &= (18 - 9) - (0 - 0) \ &= 9 - 0 \ &= 9 \end{align} $$

$$ \begin{align} \int^5_1 (-2x^3 + 3x^2 + x -5)dx &= \left[- \frac{1}{2}x^4 + x^3 + \frac{1}{2}x^2 -5x\right]^5_1 \ &= (-\frac{625}{2} + 125 + \frac{25}{2} -25) - (-\frac{1}{2} + 1 + \frac{1}{2} -5) \ &= (-300 + 100) - (-4) \ &= -196 \ \end{align} $$

$$ \begin{align} \int^5_1 (12x^5 - 5x^4 + 8x^3 -9x^2 -4x + 3)dx \end{align} $$