以下の定積分、不定積分を計算します。 (4) $\int 4x^7 dx$ (5) $\int (x+3)(x-3) dx$ (6) $\int_{-1}^{1} 2x^4 dx$ (7) $\int_{0}^{1} (4x^3 + 3x^2 + 2x + 1) dx$ (8) $\int_{0}^{1} (4x - 3) dx$ (9) $\int_{1}^{2} (x^2 - x) dx$ (10) $\int_{-1}^{1} (x+\frac{1}{2})^2 dx$

はい、承知いたしました。画像に写っている積分問題を解きます。

1. 問題の内容

以下の定積分、不定積分を計算します。

(4) $\int 4x^7 dx$

(5) $\int (x+3)(x-3) dx$

(6) $\int_{-1}^{1} 2x^4 dx$

(7) $\int_{0}^{1} (4x^3 + 3x^2 + 2x + 1) dx$

(8) $\int_{0}^{1} (4x - 3) dx$

(9) $\int_{1}^{2} (x^2 - x) dx$

(10) $\int_{-1}^{1} (x+\frac{1}{2})^2 dx$

2. 解き方の手順

(4)

$\int 4x^7 dx = 4 \int x^7 dx = 4 \cdot \frac{x^8}{8} + C = \frac{1}{2}x^8 + C$

(5)

$\int (x+3)(x-3) dx = \int (x^2 - 9) dx = \int x^2 dx - \int 9 dx = \frac{x^3}{3} - 9x + C$

(6)

$\int_{-1}^{1} 2x^4 dx = 2 \int_{-1}^{1} x^4 dx = 2 \left[ \frac{x^5}{5} \right]_{-1}^{1} = 2 \left( \frac{1^5}{5} - \frac{(-1)^5}{5} \right) = 2 \left( \frac{1}{5} - \frac{-1}{5} \right) = 2 \cdot \frac{2}{5} = \frac{4}{5}$

(7)

$\int_{0}^{1} (4x^3 + 3x^2 + 2x + 1) dx = \left[ x^4 + x^3 + x^2 + x \right]_{0}^{1} = (1^4 + 1^3 + 1^2 + 1) - (0^4 + 0^3 + 0^2 + 0) = 1 + 1 + 1 + 1 = 4$

(8)

$\int_{0}^{1} (4x - 3) dx = \left[ 2x^2 - 3x \right]_{0}^{1} = (2(1)^2 - 3(1)) - (2(0)^2 - 3(0)) = 2 - 3 = -1$

(9)

$\int_{1}^{2} (x^2 - x) dx = \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_{1}^{2} = \left( \frac{2^3}{3} - \frac{2^2}{2} \right) - \left( \frac{1^3}{3} - \frac{1^2}{2} \right) = \left( \frac{8}{3} - 2 \right) - \left( \frac{1}{3} - \frac{1}{2} \right) = \frac{8}{3} - 2 - \frac{1}{3} + \frac{1}{2} = \frac{7}{3} - \frac{3}{2} = \frac{14 - 9}{6} = \frac{5}{6}$

(10)

$\int_{-1}^{1} (x+\frac{1}{2})^2 dx = \int_{-1}^{1} (x^2 + x + \frac{1}{4}) dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} + \frac{1}{4}x \right]_{-1}^{1} = \left( \frac{1^3}{3} + \frac{1^2}{2} + \frac{1}{4}(1) \right) - \left( \frac{(-1)^3}{3} + \frac{(-1)^2}{2} + \frac{1}{4}(-1) \right) = \left( \frac{1}{3} + \frac{1}{2} + \frac{1}{4} \right) - \left( -\frac{1}{3} + \frac{1}{2} - \frac{1}{4} \right) = \frac{1}{3} + \frac{1}{2} + \frac{1}{4} + \frac{1}{3} - \frac{1}{2} + \frac{1}{4} = \frac{2}{3} + \frac{2}{4} = \frac{2}{3} + \frac{1}{2} = \frac{4 + 3}{6} = \frac{7}{6}$

3. 最終的な答え

(4) $\frac{1}{2}x^8 + C$

(5) $\frac{x^3}{3} - 9x + C$

(6) $\frac{4}{5}$

(7) $4$

(8) $-1$

(9) $\frac{5}{6}$

(10) $\frac{7}{6}$