与えられた積分を計算する問題です。積分定数を$C$とします。 (1) $\int (x^2 + \frac{3}{2}x + 2) \, dx$ (2) $\int x^3 (3x - 2) \, dx$ (3) $\int_{0}^{2} (-2x + 5) \, dx$ (4) $\int_{1}^{-3} (x^2 + 3) \, dx - \int_{2}^{-3} (x^2 + 3) \, dx$ (5) $\int_{-3}^{2} (x+3)(x-2) \, dx$ (6) $\int_{0}^{3} |x^2 - 1| \, dx$

1. 問題の内容

与えられた積分を計算する問題です。積分定数を$C$とします。

(1) $\int (x^2 + \frac{3}{2}x + 2) \, dx$

(2) $\int x^3 (3x - 2) \, dx$

(3) $\int_{0}^{2} (-2x + 5) \, dx$

(4) $\int_{1}^{-3} (x^2 + 3) \, dx - \int_{2}^{-3} (x^2 + 3) \, dx$

(5) $\int_{-3}^{2} (x+3)(x-2) \, dx$

(6) $\int_{0}^{3} |x^2 - 1| \, dx$

2. 解き方の手順

(1)

$\int (x^2 + \frac{3}{2}x + 2) \, dx = \int x^2 \, dx + \frac{3}{2} \int x \, dx + 2 \int 1 \, dx$

$= \frac{1}{3}x^3 + \frac{3}{2} \cdot \frac{1}{2}x^2 + 2x + C$

$= \frac{1}{3}x^3 + \frac{3}{4}x^2 + 2x + C$

(2)

$\int x^3 (3x - 2) \, dx = \int (3x^4 - 2x^3) \, dx = 3 \int x^4 \, dx - 2 \int x^3 \, dx$

$= 3 \cdot \frac{1}{5}x^5 - 2 \cdot \frac{1}{4}x^4 + C$

$= \frac{3}{5}x^5 - \frac{1}{2}x^4 + C$

(3)

$\int_{0}^{2} (-2x + 5) \, dx = [-x^2 + 5x]_{0}^{2} = (-2^2 + 5 \cdot 2) - (-0^2 + 5 \cdot 0)$

$= -4 + 10 = 6$

(4)

$\int_{1}^{-3} (x^2 + 3) \, dx - \int_{2}^{-3} (x^2 + 3) \, dx = \int_{1}^{-3} (x^2 + 3) \, dx + \int_{-3}^{2} (x^2 + 3) \, dx = \int_{1}^{2} (x^2 + 3) \, dx$

$= [\frac{1}{3}x^3 + 3x]_{1}^{2} = (\frac{1}{3} \cdot 2^3 + 3 \cdot 2) - (\frac{1}{3} \cdot 1^3 + 3 \cdot 1) = \frac{8}{3} + 6 - \frac{1}{3} - 3 = \frac{7}{3} + 3 = \frac{7+9}{3} = \frac{16}{3}$

(5)

$\int_{-3}^{2} (x+3)(x-2) \, dx = \int_{-3}^{2} (x^2 + x - 6) \, dx = [\frac{1}{3}x^3 + \frac{1}{2}x^2 - 6x]_{-3}^{2}$

$= (\frac{1}{3} \cdot 2^3 + \frac{1}{2} \cdot 2^2 - 6 \cdot 2) - (\frac{1}{3} \cdot (-3)^3 + \frac{1}{2} \cdot (-3)^2 - 6 \cdot (-3))$

$= (\frac{8}{3} + 2 - 12) - (-9 + \frac{9}{2} + 18) = \frac{8}{3} - 10 - (9 + \frac{9}{2}) = \frac{8}{3} - 10 - 9 - \frac{9}{2} = \frac{16}{6} - \frac{60}{6} - \frac{54}{6} - \frac{27}{6} = \frac{16-60-54-27}{6} = \frac{-125}{6}$

(6)

$\int_{0}^{3} |x^2 - 1| \, dx = \int_{0}^{1} (1 - x^2) \, dx + \int_{1}^{3} (x^2 - 1) \, dx$

$= [x - \frac{1}{3}x^3]_{0}^{1} + [\frac{1}{3}x^3 - x]_{1}^{3} = (1 - \frac{1}{3}) - (0 - 0) + (\frac{1}{3} \cdot 3^3 - 3) - (\frac{1}{3} \cdot 1^3 - 1)$

$= \frac{2}{3} + (9 - 3) - (\frac{1}{3} - 1) = \frac{2}{3} + 6 - (-\frac{2}{3}) = \frac{2}{3} + 6 + \frac{2}{3} = \frac{4}{3} + 6 = \frac{4+18}{3} = \frac{22}{3}$

3. 最終的な答え

(1) $\frac{1}{3}x^3 + \frac{3}{4}x^2 + 2x + C$

(2) $\frac{3}{5}x^5 - \frac{1}{2}x^4 + C$

(3) $6$

(4) $\frac{16}{3}$

(5) $-\frac{125}{6}$

(6) $\frac{22}{3}$