与えられた定積分の和を計算します。つまり、 $\int_{-1}^{2} (9x^2 + 8x + 7) \, dx + \int_{2}^{3} (9x^2 + 8x + 7) \, dx$ を計算します。解析学定積分積分不定積分2025/4/61. 問題の内容与えられた定積分の和を計算します。つまり、∫−12(9x2+8x+7) dx+∫23(9x2+8x+7) dx\int_{-1}^{2} (9x^2 + 8x + 7) \, dx + \int_{2}^{3} (9x^2 + 8x + 7) \, dx∫−12(9x2+8x+7)dx+∫23(9x2+8x+7)dxを計算します。2. 解き方の手順まず、不定積分を計算します。∫(9x2+8x+7) dx=9∫x2 dx+8∫x dx+7∫1 dx=9⋅x33+8⋅x22+7x+C=3x3+4x2+7x+C\int (9x^2 + 8x + 7) \, dx = 9 \int x^2 \, dx + 8 \int x \, dx + 7 \int 1 \, dx = 9 \cdot \frac{x^3}{3} + 8 \cdot \frac{x^2}{2} + 7x + C = 3x^3 + 4x^2 + 7x + C∫(9x2+8x+7)dx=9∫x2dx+8∫xdx+7∫1dx=9⋅3x3+8⋅2x2+7x+C=3x3+4x2+7x+C次に、定積分を計算します。∫−12(9x2+8x+7) dx=[3x3+4x2+7x]−12=(3(2)3+4(2)2+7(2))−(3(−1)3+4(−1)2+7(−1))=(3(8)+4(4)+14)−(3(−1)+4(1)−7)=(24+16+14)−(−3+4−7)=(54)−(−6)=54+6=60\int_{-1}^{2} (9x^2 + 8x + 7) \, dx = [3x^3 + 4x^2 + 7x]_{-1}^{2} = (3(2)^3 + 4(2)^2 + 7(2)) - (3(-1)^3 + 4(-1)^2 + 7(-1)) = (3(8) + 4(4) + 14) - (3(-1) + 4(1) - 7) = (24 + 16 + 14) - (-3 + 4 - 7) = (54) - (-6) = 54 + 6 = 60∫−12(9x2+8x+7)dx=[3x3+4x2+7x]−12=(3(2)3+4(2)2+7(2))−(3(−1)3+4(−1)2+7(−1))=(3(8)+4(4)+14)−(3(−1)+4(1)−7)=(24+16+14)−(−3+4−7)=(54)−(−6)=54+6=60∫23(9x2+8x+7) dx=[3x3+4x2+7x]23=(3(3)3+4(3)2+7(3))−(3(2)3+4(2)2+7(2))=(3(27)+4(9)+21)−(3(8)+4(4)+14)=(81+36+21)−(24+16+14)=(138)−(54)=84\int_{2}^{3} (9x^2 + 8x + 7) \, dx = [3x^3 + 4x^2 + 7x]_{2}^{3} = (3(3)^3 + 4(3)^2 + 7(3)) - (3(2)^3 + 4(2)^2 + 7(2)) = (3(27) + 4(9) + 21) - (3(8) + 4(4) + 14) = (81 + 36 + 21) - (24 + 16 + 14) = (138) - (54) = 84∫23(9x2+8x+7)dx=[3x3+4x2+7x]23=(3(3)3+4(3)2+7(3))−(3(2)3+4(2)2+7(2))=(3(27)+4(9)+21)−(3(8)+4(4)+14)=(81+36+21)−(24+16+14)=(138)−(54)=84最後に、定積分の和を計算します。∫−12(9x2+8x+7) dx+∫23(9x2+8x+7) dx=60+84=144\int_{-1}^{2} (9x^2 + 8x + 7) \, dx + \int_{2}^{3} (9x^2 + 8x + 7) \, dx = 60 + 84 = 144∫−12(9x2+8x+7)dx+∫23(9x2+8x+7)dx=60+84=144別の解法:積分区間がつながっているので、∫−12(9x2+8x+7) dx+∫23(9x2+8x+7) dx=∫−13(9x2+8x+7) dx\int_{-1}^{2} (9x^2 + 8x + 7) \, dx + \int_{2}^{3} (9x^2 + 8x + 7) \, dx = \int_{-1}^{3} (9x^2 + 8x + 7) \, dx∫−12(9x2+8x+7)dx+∫23(9x2+8x+7)dx=∫−13(9x2+8x+7)dx∫−13(9x2+8x+7) dx=[3x3+4x2+7x]−13=(3(3)3+4(3)2+7(3))−(3(−1)3+4(−1)2+7(−1))=(3(27)+4(9)+21)−(3(−1)+4(1)−7)=(81+36+21)−(−3+4−7)=(138)−(−6)=138+6=144\int_{-1}^{3} (9x^2 + 8x + 7) \, dx = [3x^3 + 4x^2 + 7x]_{-1}^{3} = (3(3)^3 + 4(3)^2 + 7(3)) - (3(-1)^3 + 4(-1)^2 + 7(-1)) = (3(27) + 4(9) + 21) - (3(-1) + 4(1) - 7) = (81 + 36 + 21) - (-3 + 4 - 7) = (138) - (-6) = 138 + 6 = 144∫−13(9x2+8x+7)dx=[3x3+4x2+7x]−13=(3(3)3+4(3)2+7(3))−(3(−1)3+4(−1)2+7(−1))=(3(27)+4(9)+21)−(3(−1)+4(1)−7)=(81+36+21)−(−3+4−7)=(138)−(−6)=138+6=1443. 最終的な答え144